Convex Hull
William Elliot
Are convex sets defined for vector spaces with complex scalars?
Jimmy John
William Elliot <ma...@rdrop.remove.com> wrote:
> Is the convex hull of an open set, open?
>
> Are convex sets defined for vector spaces with complex scalars?
For a real space the convex hull of a set S is such that every point
"between" two points of the set is also in the set.
Thus if x and y are members of the convex hul of S, and 0 < a < 1 so are
all of a*x + (1-a)*y in that hull.
BUt 0 < a < 1 does not work as nicely for complex scalars, a.
--
Science is based directly on objective physical evidence,
and nothing that is not based directly on objective physical evidence
can be science.
David C. Ullrich
<ma...@rdrop.remove.com> wrote:
>Is the convex hull of an open set, open?
Please post your questions using correct English grammar.
Those, extra, commas, make, it, harder, to, parse, the, question.
You don't specify the context. It's extremely easy to show
just from the definitions that the convex hull of an open
set in a topological vector space (meaning real or complex
topological vector space, as in most common) is open.
>Are convex sets defined for vector spaces with complex scalars?
Of course they are. The definition is the same as for real
vector spaces.
Robert Israel
> In article <2011030519...@agora.rdrop.com>,
> William Elliot <ma...@rdrop.remove.com> wrote:
>
>> Is the convex hull of an open set, open?
>>
>> Are convex sets defined for vector spaces with complex scalars?
>
> For a real space the convex hull of a set S is such that every point
> "between" two points of the set is also in the set.
>
> Thus if x and y are members of the convex hul of S, and 0 < a < 1 so are
> all of a*x + (1-a)*y in that hull.
>
>
> BUt 0 < a < 1 does not work as nicely for complex scalars, a.
Nonsense. If 0 < a < 1 then "a" is a real number. The vector space is
over the complex numbers, but some of the scalars are real.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
William Elliot
> <ma...@rdrop.remove.com> wrote:
>
>> Is the convex hull of an open set, open?
> You don't specify the context. It's extremely easy to show
> just from the definitions that the convex hull of an open
> set in a topological vector space (meaning real or complex
> topological vector space, as in most common) is open.
>
Ok, I think I get it. The notion of convex relies upon the field
being (partially) ordered. If it's trivially ordered, then every
set is it's own convex hull. If the field isn't trivially ordered,
then 0 < 1 and the prime field is orderfield isomorphic to Q.
For the complex field C, which can't be linearly ordered, we have
various non-trivially choices for ordering C. For example using a
positive cone of positive rationals or of positive reals.
For a real vector space with R ordered by the positive cone of positive
rationals, the convex hull of an open set may not be open. On the other
hand, were R totally ordered, then I think the convex hull of an open
set is open. Alas, a proof of that eludes me.
>> Are convex sets defined for vector spaces with complex scalars?
> Of course they are. The definition is the same as for real
> vector spaces.
As for a complex vector space, I'm not so sure convex hulls of open sets
are open. Likely that's because complex analysis hasn't been made simple.
Here's what I'm puzzle about. To simplify the discussion,
let the questions below be for real topological vector spaces
with the topology of the scalars R, coincident with the linear
order topology of the usual total order of R.
If U is an open set of a topological vector space, is the
convex hull of U open?
If U is closed set of a topological vector space, is the
convex hull of U closed?
If U is a convex set, is the closure of U convex?
If U is a convex set and U subset K subset cl U, is K convex?
If the vector space has finite dimension, will that change any
of the answers?
----
achille
>
> Here's what I'm puzzle about. To simplify the discussion,
> let the questions below be for real topological vector spaces
> with the topology of the scalars R, coincident with the linear
> order topology of the usual total order of R.
>
> If U is an open set of a topological vector space, is the
> convex hull of U open?
true: obvious
> If U is closed set of a topological vector space, is the
> convex hull of U closed?
false: U = { (x,1/(1+x^2)) : x in R }
> If U is a convex set, is the closure of U convex?
true: U convex => both int U and cl U convex.
> If U is a convex set and U subset K subset cl U, is K convex?
false: U = open unit square, K = U plus U's corners
William Elliot
> On Mar 7, 12:12 pm, William Elliot <ma...@rdrop.remove.com> wrote:
>>
>> Here's what I'm puzzle about. To simplify the discussion,
>> let the questions below be for real topological vector spaces
>> with the topology of the scalars R, coincident with the linear
>> order topology of the usual total order of R.
>>
>> If U is an open set of a topological vector space, is the
>> convex hull of U open?
>
> true: obvious
>
U subset convex hull U
U = int U subset int convex hull U
convex hull U subset int convex hull U
. . by definition of convex hull and because int convex hull U is convex
convex hull U = int convex hull U
BTW, what notation is used for convex hull?
>> If U is closed set of a topological vector space, is the
>> convex hull of U closed?
>
> false: U = { (x,1/(1+x^2)) : x in R }
>
Is it true if U is topologically bounded?
>> If U is a convex set, is the closure of U convex?
>
> true: U convex => both int U and cl U convex.
What steps are needed to prove those?
William Elliot
for all r,s > 0, (r + s)U = rU + sU.
Now since non-zero scalar multiplication is a homeomorphism
and int A + int B subset int A+B, we have
r.int U + s.int U = int rU + int sU
subset int (rU + sU) = int (r + s)U
= (r + s).int U subset r.int U + s.int U
(r + s).int U = r.int U + s.int U and int U is convex.
Likewise:
r.cl U + s.cl U = cl rU + cl sU
subset cl (rU + sU) = cl (r + s)U
= (r + s).cl U subset r.cl U + s.cl U
(r + s).cl U = r.cl U + s.cl U and cl U is convex.
----
David C. Ullrich
<ma...@rdrop.remove.com> wrote:
>On Sun, 6 Mar 2011, David C. Ullrich wrote:
>> <ma...@rdrop.remove.com> wrote:
>>
>>> Is the convex hull of an open set, open?
>
>> You don't specify the context. It's extremely easy to show
>> just from the definitions that the convex hull of an open
>> set in a topological vector space (meaning real or complex
>> topological vector space, as in most common) is open.
>>
>Ok, I think I get it. The notion of convex relies upon the field
>being (partially) ordered. If it's trivially ordered, then every
>set is it's own convex hull. If the field isn't trivially ordered,
>then 0 < 1 and the prime field is orderfield isomorphic to Q.
>
>For the complex field C, which can't be linearly ordered, we have
>various non-trivially choices for ordering C. For example using a
>positive cone of positive rationals or of positive reals.
>
>For a real vector space with R ordered by the positive cone of positive
>rationals, the convex hull of an open set may not be open. On the other
>hand, were R totally ordered, then I think the convex hull of an open
>set is open. Alas, a proof of that eludes me.
Whatever. Why not just look up the definition? If S is a subset
of a real or complex vector space then S is convex if
x, y in S, 0 <= t <= 1 implies tx + (1-t)y in S. Period.
That's a perfectly standard definition.
>>> Are convex sets defined for vector spaces with complex scalars?
>
>> Of course they are. The definition is the same as for real
>> vector spaces.
>
>As for a complex vector space, I'm not so sure convex hulls of open sets
>are open. Likely that's because complex analysis hasn't been made simple.
>
>Here's what I'm puzzle about. To simplify the discussion,
>let the questions below be for real topological vector spaces
>with the topology of the scalars R, coincident with the linear
>order topology of the usual total order of R.
>
>If U is an open set of a topological vector space, is the
>convex hull of U open?
Yes. The proof is trivial. Hint: The vector-space operations
are continuous.
Ok, now you say you don't see how to prove it. Fine.
Lemma. If A is open then A + B = {a + b : a in A, b in B}
is open.
Pf: The set b + A is open for every b in A, and A + B
is the union of these sets. QED.
Now note that if O is open then the convex hull of
O is the union of O and the sets
tO + (1-t)O
for 0 < t < 1.
>If U is closed set of a topological vector space, is the
>convex hull of U closed?
No.
>If U is a convex set, is the closure of U convex?
Yes, this is trivial.
>If U is a convex set and U subset K subset cl U, is K convex?
No. You can easily construct an example in the plane.
With U an open square, for example.
(This is why "fuzzy balls" don't have to be convex...)
>
>If the vector space has finite dimension, will that change any
>of the answers?
Not as far as I can see. You might note however that the
convex hull of a _compact_ set is compact. (Hint:
If K is a compact subset of the TVS X, let S = ___
and define F : S -> X by ___. The Tychonoff theorem
shows that S is compact; now F is continuous and
the convex hull of K is F(S), QED).
>
>----
achille
> On Sun, 6 Mar 2011, achille wrote:
>
> >> Here's what I'm puzzle about. �To simplify the discussion,
> >> let the questions below be for real topological vector spaces
> >> with the topology of the scalars R, coincident with the linear
> >> order topology of the usual total order of R.
>
> >> If U is an open set of a topological vector space, is the
> >> convex hull of U open?
>
> > true: obvious
>
> Obvious? It's simple if interiors of convex sets are convex.
> U subset convex hull U
> U = int U subset int convex hull U
> convex hull U subset int convex hull U
> . . by definition of convex hull and because int convex hull U is convex
> convex hull U = int convex hull U
>
> BTW, what notation is used for convex hull?
>
> >> If U is closed set of a topological vector space, is the
> >> convex hull of U closed?
>
> > false: U = { (x,1/(1+x^2)) : x in R }
>
> Is it true if U is topologically bounded?
>
"True" if U is a bounded and closed subset of R^n
for finite n. It is pretty easy to prove this using
Carathéodory's theorem:
For A \subset R^n, every point in conv A is a
convex combination of <= n+1 points from A.
No idea what happens when n is infinite.
achille
>
>
> >If the vector space has finite dimension, will that change any
> >of the answers?
>
> Not as far as I can see. You might note however that the
> convex hull of a _compact_ set is compact. (Hint:
> If K is a compact subset of the TVS X, let S = ___
> and define F : S -> X by ___. The Tychonoff theorem
> shows that S is compact; now F is continuous and
> the convex hull of K is F(S), QED).
>
Hmm, a web search indicate this is false for infinite
dimensional TVS X.
Counterexample: Consider l_2, the space of all square
summable sequences. For each n, let u_n be the sequence:
(u_n)_i = 1/n if i = n,
= 0 if i != n.
Let A be the set {u_1, u_2, .. } U {0}, then A is a norm
compact subset of l_2 but conv A is not even closed.
Am I missing something?
REF: Charalambos D. Aliprantis, Kim C. Border,
Infinite Dimensional Analysis: A Hitchhiker's Guide
Example 5.34 (Noncompact convex hull) on page 185.
URL:
http://books.google.com/books?id=4hIq6ExH7NoC&lpg=PP1&dq=Infinite%20Dimensional%20Analysis%3A%20A%20Hitchhiker's%20Guide&pg=PA185#v=onepage&q=Noncompact%20convex%20hull&f=false
Robert Israel
False. For example, in a separable Hilbert space let U be an
orthonormal basis {u_1, u_2, ...}. This is closed since
it is discrete: ||u_i - u_j|| = sqrt(2) for all i <> j.
Then conv U is the set of vectors sum_i c_i u_i with all but
finitely many c_i = 0, all c_i >= 0, and sum_i c_i = 1.
But that is not closed: e.g. sum_{i=1}^infty 2^(-i) u_i is
in its closure but is not in conv U.
Ilmari Karonen
> On Sun, 6 Mar 2011 20:12:22 -0800, William Elliot
><ma...@rdrop.remove.com> wrote:
>
>>If U is a convex set and U subset K subset cl U, is K convex?
>
> No. You can easily construct an example in the plane.
> With U an open square, for example.
>
> (This is why "fuzzy balls" don't have to be convex...)
Hmm... so what conditions does one need for fuzzy balls to be convex,
then?
An obviously necessary and sufficient condition is that the boundary
of a ball contains no three collinear points, but that is little more
than a restatement of the question.
It seems to me that the norm needs to be somehow degenerate in order
to allow non-convex fuzzy balls, but I'm having a hard time recalling
enough of this stuff to formalize it properly.
--
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.
William Elliot
>>>> convex hull of U closed?
>>
>>> false: U = { (x,1/(1+x^2)) : x in R }
>>
>> Is it true if U is topologically bounded?
>
> "True" if U is a bounded and closed subset of R^n
> for finite n. It is pretty easy to prove this using
> Carathéodory's theorem:
>
> For A \subset R^n, every point in conv A is a
> convex combination of <= n+1 points from A.
>
An interesting counter example is the closed set {0,1} of the one
dimensional vector space of R^1 with the rationals as scalars.
William Elliot
>> convex hull of U open?
>
> Yes. The proof is trivial. Hint: The vector-space operations
> are continuous.
>
> Ok, now you say you don't see how to prove it. Fine.
What do you think of the proof I gave of that, in the other subtread,
along with showing both the interior and the closure of a convex set is
convex?
> Lemma. If A is open then A + B = {a + b : a in A, b in B}
> is open.
>
> Pf: The set b + A is open for every b in A, and A + B
> is the union of these sets. QED.
>
> Now note that if O is open then the convex hull of
> O is the union of O and the sets
>
> tO + (1-t)O
>
> for 0 < t < 1.
I doubt that \/{ tO + (1 - 1)O | t in [0,1] } is the
convex hull of O. For example if O = { (0,0), (0,1), (1,0) }
and for an open example, the same with an open ball of
radius 10^-100 centered at each point instead.
>> If U is a convex set, is the closure of U convex?
>
> Yes, this is trivial.
>
Trivial? My proof was straight forward.
Have you a proof that's trivially easy?
> You might note however that the convex hull of a _compact_ set is
> compact.
Indeed, an interesting problem. Seems it has limited application.
For a counter example, {0,1} of the vector space R^1 with scalars Q.
> (Hint: If K is a compact subset of the TVS X, let S = ___ and
> define F : S -> X by ___. The Tychonoff theorem shows that S is compact;
> now F is continuous and the convex hull of K is F(S), QED).
What's the Tychonoff theorem?
The product of compact spaces is compact?
----
David C. Ullrich
<achil...@yahoo.com.hk> wrote:
>On Mar 7, 11:55 pm, David C. Ullrich <ullr...@math.okstate.edu> wrote:
>>
>>
>> >If the vector space has finite dimension, will that change any
>> >of the answers?
>>
>> Not as far as I can see. You might note however that the
>> convex hull of a _compact_ set is compact. (Hint:
>> If K is a compact subset of the TVS X, let S = ___
>> and define F : S -> X by ___. The Tychonoff theorem
>> shows that S is compact; now F is continuous and
>> the convex hull of K is F(S), QED).
>>
>
>Hmm, a web search indicate this is false for infinite
>dimensional TVS X.
Yeah, realized this morning that I lied, sorry.
I also lied about something else, saying that the convex
hull of O was the set of convex combinations of
_pairs_ of elements of O. There's a connection
between the two lies...
David C. Ullrich
If the norm is "strictly convex" then fuzzy balls are convex.
Of course if you look at the definition you see that that's also
just saying that the boundary contains no stright line segments...
David C. Ullrich
<ma...@rdrop.remove.com> wrote:
>>> If U is an open set of a topological vector space, is the
>>> convex hull of U open?
>>
>> Yes. The proof is trivial. Hint: The vector-space operations
>> are continuous.
>>
>> Ok, now you say you don't see how to prove it. Fine.
>
>What do you think of the proof I gave of that, in the other subtread,
>along with showing both the interior and the closure of a convex set is
>convex?
>
>> Lemma. If A is open then A + B = {a + b : a in A, b in B}
>> is open.
>>
>> Pf: The set b + A is open for every b in A, and A + B
>> is the union of these sets. QED.
>>
>> Now note that if O is open then the convex hull of
>> O is the union of O and the sets
>>
>> tO + (1-t)O
>>
>> for 0 < t < 1.
>
>I doubt that \/{ tO + (1 - 1)O | t in [0,1] } is the
>convex hull of O.
Yeah, I lied about that, sorry.
It's still immediate from the Lemma that the convex
hull of an open set is open. Say O is open and x is in
the convex hull of O. There exist x_1, ... x_n in
O and t_j >= 0 with sum t_j = 1 and sum t_j x_j = x.
Not all the t_j vanish. Say t_1 <> 0. Let
S = t_1 O + sum_{j>1} t_j x_j.
The lemma shows that S is open, and it's clear that
x is in S and S is a subset of the convex hull of O.
> For example if O = { (0,0), (0,1), (1,0) }
>and for an open example, the same with an open ball of
>radius 10^-100 centered at each point instead.
>
>>> If U is a convex set, is the closure of U convex?
>>
>> Yes, this is trivial.
>>
>Trivial? My proof was straight forward.
>Have you a proof that's trivially easy?
As often happens, things are most obvious if we think
in terms of nets. Say C is convex, K is the closure of C,
x , y are in K and 0 < t < 1.
There exist nets x_j and y_j in C with x_j -> x and
y_j -> y. Continuity of the vector-space operations
shows that t x_j + (1-t) y_j -> tx + (1-t)y; hence
tx + (1-t)y is in K.
(Hmm, can you spot the slight technicality I'm
overlooking there and fix it?)
>> You might note however that the convex hull of a _compact_ set is
>> compact.
This is false, by the way. Sorry.
>Indeed, an interesting problem. Seems it has limited application.
>For a counter example, {0,1} of the vector space R^1 with scalars Q.
Here you're getting back to your own private use of the word
"convex"...
>
>> (Hint: If K is a compact subset of the TVS X, let S = ___ and
>> define F : S -> X by ___. The Tychonoff theorem shows that S is compact;
>> now F is continuous and the convex hull of K is F(S), QED).
>
>What's the Tychonoff theorem?
>The product of compact spaces is compact?
Yes.
>
>----
Rob Johnson
Robert Israel <isr...@math.MyUniversitysInitials.ca> wrote:
>On Mon, 7 Mar 2011 07:59:09 -0800 (PST), achille wrote:
>
>> On Mar 7, 6:01 pm, William Elliot <ma...@rdrop.remove.com> wrote:
>>> On Sun, 6 Mar 2011, achille wrote:
>>>
>>>>> Here's what I'm puzzle about. To simplify the discussion,
>>>>> let the questions below be for real topological vector spaces
>>>>> with the topology of the scalars R, coincident with the linear
>>>>> order topology of the usual total order of R.
>>>
>>>>> If U is an open set of a topological vector space, is the
>>>>> convex hull of U open?
>>>
>>>> true: obvious
>>>
>>> Obvious? It's simple if interiors of convex sets are convex.
>>> U subset convex hull U
>>> U = int U subset int convex hull U
>>> convex hull U subset int convex hull U
>>> . . by definition of convex hull and because int convex hull U is convex
>>> convex hull U = int convex hull U
>>>
>>> BTW, what notation is used for convex hull?
>>>
>>>>> If U is closed set of a topological vector space, is the
>>>>> convex hull of U closed?
>>>
>>>> false: U = { (x,1/(1+x^2)) : x in R }
>>>
>>> Is it true if U is topologically bounded?
>>>
>>
>> "True" if U is a bounded and closed subset of R^n
>> for finite n. It is pretty easy to prove this using
>>
>> For A \subset R^n, every point in conv A is a
>> convex combination of <= n+1 points from A.
>>
>> No idea what happens when n is infinite.
>
>False. For example, in a separable Hilbert space let U be an
>orthonormal basis {u_1, u_2, ...}. This is closed since
>it is discrete: ||u_i - u_j|| = sqrt(2) for all i <> j.
>Then conv U is the set of vectors sum_i c_i u_i with all but
>finitely many c_i = 0, all c_i >= 0, and sum_i c_i = 1.
>But that is not closed: e.g. sum_{i=1}^infty 2^(-i) u_i is
>in its closure but is not in conv U.
A Hilbert space is an inner product space which is a vector space.
Now, the elements of a vector space are finite linear combinations
of the basis vectors. However, in dealing with Fourier series, the
set of functions U = { exp(i2pi k x) : k in Z } is a basis for the
Hilbert space L^2(R/Z). However, a Hilbert space is complete, and
so the function
oo
--- -k
f(x) = > 2 cos(2pi k x)
---
k=1
is in L^2(R/Z). However, f(x) cannot be written as a finite linear
combination of functions in U. This simply shows that the basis for
L^2(R/Z) as a vector space is different than its basis as a Hilbert
space.
Furthermore, f is also in any convex set that contains U, so it
should be in Conv(U). Thus, the elements of the convex hull in a
Hilbert space do not seem to be finite linear combinations of the
Hilbert basis elements. Your result may be correct in the vector
space setting, but it seems that the example needs some reworking
(or my understanding of things is in need of reworking, which is
quite possible).
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
to view any ASCII art, display article in a monospaced font
Rob Johnson
William Elliot <ma...@rdrop.remove.com> wrote:
>>>>> If U is closed set of a topological vector space, is the
>>>>> convex hull of U closed?
>>>
>>>> false: U = { (x,1/(1+x^2)) : x in R }
>>>
>>> Is it true if U is topologically bounded?
>>
>> "True" if U is a bounded and closed subset of R^n
>> for finite n. It is pretty easy to prove this using
>>
>> For A \subset R^n, every point in conv A is a
>> convex combination of <= n+1 points from A.
>>
>
>An interesting counter example is the closed set {0,1} of the one
>dimensional vector space of R^1 with the rationals as scalars.
Perhaps I am misreading your example, but it appears that you are
referring to the Hamel basis of R over Q
<http://mathworld.wolfram.com/HamelBasis.html>
In which case, R is (uncountably) infinite dimensional over Q.
William Elliot
> <ma...@rdrop.remove.com> wrote:
>
>>>> If U is an open set of a topological vector space, is the
>>>> convex hull of U open?
>>>
>>> is open.
>>>
> hull of an open set is open. Say O is open and x is in
> the convex hull of O. There exist x_1, ... x_n in
> O and t_j >= 0 with sum t_j = 1 and sum t_j x_j = x.
>
What if the space is R^N and the set O = { zj | j in N }
where zj(n) = 0 if n /= j and zj(j) = 1?
What if the space is R^R and O = { f_r | r in R } where
f_r(x) = 0 if x /= r and f_r(r) = 1?
> Not all the t_j vanish. Say t_1 <> 0. Let
> S = t_1 O + sum_{j>1} t_j x_j.
>
> The lemma shows that S is open, and it's clear that
> x is in S and S is a subset of the convex hull of O.
>
It's not clear that S subset convex hull O.
On the other hand, my proof given previously in the other subthread,
is short and lacks none of those loose ends. Have you looked at it?
>> For example if O = { (0,0), (0,1), (1,0) }
>> and for an open example, the same with an open ball of
>> radius 10^-100 centered at each point instead.
>>
>>>> If U is a convex set, is the closure of U convex?
>>> Yes, this is trivial.
>> Trivial? My proof was straight forward.
>> Have you a proof that's trivially easy?
>
> As often happens, things are most obvious if we think
> in terms of nets. Say C is convex, K is the closure of C,
> x , y are in K and 0 < t < 1.
>
> There exist nets x_j and y_j in C with x_j -> x and
> y_j -> y. Continuity of the vector-space operations
> shows that t x_j + (1-t) y_j -> tx + (1-t)y; hence
> tx + (1-t)y is in K.
>
> (Hmm, can you spot the slight technicality I'm
> overlooking there and fix it?)
For all y, x_j + (1 - t)y_j in C.
----
David C. Ullrich
This almost sounds like you're saying that the way a basis in a
Hilbbert space works is not the same as the way a basis in
other vector spaces work. Of course that's not so; the point
is just that the _word_ "basis" is used in different ways in
different contexts. If we called the two concepts "basis_1"
and "basis_2" to clarify things, then the way a basis_1
works in a Hilbert space is identical to other vector spaces.
Of course you understand that, I point this out explicitly
just to make sense of my comment below:
>Furthermore, f is also in any convex set that contains U, so it
>should be in Conv(U).
??? This is simply not so. As far as I know, the word "convex"
is not used in two different senses; in particular I've never
seen an exposition where "convex" in a Hilbert space
means "closed convex".
More to the point, even if the word "convex" _were_
used in two different senses here, like the word
"basis", your example would still be irrelevant, because
Robert was talking about the pure-vector-space
notion of "convex" - the non-fact(?) that "convex"
sometimes means something else doesn't change the
truth of his comments.
> Thus, the elements of the convex hull in a
>Hilbert space do not seem to be finite linear combinations of the
>Hilbert basis elements. Your result may be correct in the vector
>space setting, but it seems that the example needs some reworking
>(or my understanding of things is in need of reworking, which is
>quite possible).
You seem to be assuming that "convex" means "closed
and convex" in a Hilbert space. I've never seen the word
used this way. (Not that that would actually matter.)
David C. Ullrich
<ma...@rdrop.remove.com> wrote:
>On Tue, 8 Mar 2011, David C. Ullrich wrote:
>> <ma...@rdrop.remove.com> wrote:
>>
>>>>> If U is an open set of a topological vector space, is the
>>>>> convex hull of U open?
>>>>
>>>> Lemma. If A is open then A + B = {a + b : a in A, b in B}
>>>> is open.
>>>>
>> It's still immediate from the Lemma that the convex
>> hull of an open set is open. Say O is open and x is in
>> the convex hull of O. There exist x_1, ... x_n in
>> O and t_j >= 0 with sum t_j = 1 and sum t_j x_j = x.
>>
>This stricks me as a lemma for finite dimensional spaces only.
How it stricks you doesn't seem to have much bearing on
the math. Exactly what part seems wrong?
>What if the space is R^N and the set O = { zj | j in N }
>where zj(n) = 0 if n /= j and zj(j) = 1?
What if it is? What topology are you talking about?
Is R^N a topological vector space with that topology?
Is O open in that topology? And then what makes
you think that the convex hull of O is not open?
(The only thing I can _guess_ is that you're thinking
about the product topology on R^N. But then O is
not open...)
>What if the space is R^R and O = { f_r | r in R } where
>f_r(x) = 0 if x /= r and f_r(r) = 1?
>
>> Not all the t_j vanish. Say t_1 <> 0. Let
>> S = t_1 O + sum_{j>1} t_j x_j.
>>
>> The lemma shows that S is open, and it's clear that
>> x is in S and S is a subset of the convex hull of O.
>>
>It's not clear that S subset convex hull O.
Huh??? What do you think the definition of the convex hull
of O is?
>On the other hand, my proof given previously in the other subthread,
>is short and lacks none of those loose ends.
As far as I can see you haven't pointed out any loose ends.
You've said the lemma stricks you as valid for finite-dimensional
spaces only, without explaining what the problem is,
and you say it's not clear that S is a subset of the convex
hull of O, which is just silly - it's obvious from the definition
that S is a subset of the convex hull of O.
> Have you looked at it?
>
>>> For example if O = { (0,0), (0,1), (1,0) }
>>> and for an open example, the same with an open ball of
>>> radius 10^-100 centered at each point instead.
>>>
>>>>> If U is a convex set, is the closure of U convex?
>
>>>> Yes, this is trivial.
>>> Trivial? My proof was straight forward.
>>> Have you a proof that's trivially easy?
>>
>> As often happens, things are most obvious if we think
>> in terms of nets. Say C is convex, K is the closure of C,
>> x , y are in K and 0 < t < 1.
>>
>> There exist nets x_j and y_j in C with x_j -> x and
>> y_j -> y. Continuity of the vector-space operations
>> shows that t x_j + (1-t) y_j -> tx + (1-t)y; hence
>> tx + (1-t)y is in K.
>>
>> (Hmm, can you spot the slight technicality I'm
>> overlooking there and fix it?)
>
>For all y, x_j + (1 - t)y_j in C.
No, that's not a technicality I was overlooking,
that's something that's so utterly obvious that
it didn't need to be stated. Like if I prove a
lemma about even integers and then I apply
the lemma to 2k without stating that 2k is even.
>
>----
William Elliot
> <ma...@rdrop.remove.com> wrote:
>>>
>>>>>> If U is an open set of a topological vector space, is the
>>>>>> convex hull of U open?
>>>>>
>>>>> Lemma. If A is open then A + B = {a + b : a in A, b in B}
>>>>> is open.
>>>>>
>>> It's still immediate from the Lemma that the convex
>>> hull of an open set is open.
It's immediate from the fact that the
interior of a convex set is convex.
>>> Say O is open and x is in the convex hull of O. There exist x_1, ...
>>> x_n in O and t_j >= 0 with sum t_j = 1 and sum t_j x_j = x.
Is that true of O if O is any subset or only if O is open?
>>> Not all the t_j vanish. Say t_1 <> 0. Let
>>> S = t_1 O + sum_{j>1} t_j x_j.
>>>
>>> The lemma shows that S is open, and it's clear that
>>> x is in S and S is a subset of the convex hull of O.
>>>
>> It's not clear that S subset convex hull O.
>
> Huh??? What do you think the definition of the convex hull
> of O is?
>
The smallest convex set containing O.
>>>>>> If U is a convex set, is the closure of U convex?
>>>>> Yes, this is trivial.
>>>
>>> As often happens, things are most obvious if we think
>>> in terms of nets. Say C is convex, K is the closure of C,
>>> x , y are in K and 0 < t < 1.
>>>
>>> There exist nets x_j and y_j in C with x_j -> x and
>>> y_j -> y. Continuity of the vector-space operations
>>> shows that t x_j + (1-t) y_j -> tx + (1-t)y; hence
>>> tx + (1-t)y is in K.
>>>
>>> (Hmm, can you spot the slight technicality I'm
>>> overlooking there and fix it?)
>>
>> For all y, x_j + (1 - t)y_j in C.
>
> No, that's not a technicality I was overlooking,
> that's something that's so utterly obvious that
> it didn't need to be stated.
You've assume that the two nets have the same index.
Let M be module with an ordered ring.
For example, R^2 over the integers.
What definition is used for a convex set of M?
K is convex when
for all r,s > 0, rK + sK = (r + s)K
or
for all t in [0,1], K = tK + (1 - t)K?
In vector spaces, the two are equivalent definitions of convex.
In modules, the former is stronger. Which is used for modules?
It's interesting to note that for topological modules, it makes no
difference which is used to get the theorems that the interior and
closure of convex sets are convex and the convex hulls of open sets
are open.
It does make a difference however, with the theorem:
if 0 < r <= s, 0 in A and A is convex, then rA subset sA.
Use unpack proof facility to read packed proof file (ppf).
** convex.intcl.openhull.ppf
convex A ==> int A, cl A convex. If r,s > 0: (r + s)A = rA + sA
r.int U + s.int U = int rU + int sU subset int (rU + sU), cf grouptop
r.cl U + s.cl U = cl rU + cl sU subset cl (rU + sU), cf grouptop
open U ==> convex hull U open. U subset convex hull U
U = int U subset int convex hull U; int convex hull U convex
convex hull U subset int convex hull U subset convex hull U
**
Rob Johnson
Then which of the following is not true:
1. U is a basis of the Hilbert space L^2(R/Z).
2. f(x) is in the Hilbert space L^2(R/Z).
3. f(x) cannot be written as a finite linear combination of
elements of U.
A Hilbert space is a complete inner product space, which is a vector
space with extra structure that allows limits. It is not a vector
space, since each element of a vector space is a finite linear
combination of its basis elements, whereas the elements of a Hilbert
space are limits of elements of the underlying vector space, limits
made possible by the structure added by the inner product.
U is a closed subset of the Hilbert space, but without some structure
beyond a vector space, U is simply a countable union of closed sets,
which is not necessarily closed under the standard product topology.
So we cannot say that U is closed when looked at solely as a subset
of a vector space.
>>Furthermore, f is also in any convex set that contains U, so it
>>should be in Conv(U).
>
>??? This is simply not so. As far as I know, the word "convex"
>is not used in two different senses; in particular I've never
>seen an exposition where "convex" in a Hilbert space
>means "closed convex".
>
>More to the point, even if the word "convex" _were_
>used in two different senses here, like the word
>"basis", your example would still be irrelevant, because
>Robert was talking about the pure-vector-space
>notion of "convex" - the non-fact(?) that "convex"
>sometimes means something else doesn't change the
>truth of his comments.
Robert's example uses elements and the norm of the inner product
space, then says that elements of the convex hull must be elements
of the underlying vector space by requiring all but finitely many
c_i = 0.
Now this may be where I am wrong, but in a Hilbert space, I usually
extend the idea of convexity so that a set which contains { v_k }
must also contain
oo
---
> a v
--- k k
k=1
for all a_k >= 0 and
oo
---
> a = 1
--- k
k=1
If we only require finite convex linear combinations for convexity,
then that is where I believe I am wrong.
>> Thus, the elements of the convex hull in a
>>Hilbert space do not seem to be finite linear combinations of the
>>Hilbert basis elements. Your result may be correct in the vector
>>space setting, but it seems that the example needs some reworking
>>(or my understanding of things is in need of reworking, which is
>>quite possible).
>
>You seem to be assuming that "convex" means "closed
>and convex" in a Hilbert space. I've never seen the word
>used this way. (Not that that would actually matter.)
No, I don't assume that convex means closed, but as I mentioned
above, I do extend the notion of convex in the context of a Hilbert
space, and that may be where I am wrong.
I have done a bit more research and this notion of convexity is
called sigma-convexity. So, indeed, this is where I am wrong.
<http://mathoverflow.net/questions/56161/infinite-convex-combinations-in-a-banach-space>
<http://projecteuclid.org/euclid.pjm/1102818223>
<http://projecteuclid.org/euclid.nihmj/1273779702>
<http://arxiv.org/abs/1006.1958>
David C. Ullrich
<r...@trash.whim.org> wrote:
Huh? I'm puzzled by the question, because you know as
well as I do, or I would have thought you knew as
well as I do, and I explained at some length above,
that
This depends on what you mean by "basis". There are
various definitions of the word in various contexts,
and at least two of them apply here.
>
>2. f(x) is in the Hilbert space L^2(R/Z).
>
>3. f(x) cannot be written as a finite linear combination of
> elements of U.
>
>A Hilbert space is a complete inner product space, which is a vector
>space with extra structure that allows limits. It is not a vector
>space,
A Hilbert space is not a vector space?
Huh???????????????????????????????????????
This seems like you simply didn't read what I wrote! Never mind the
fact that how _you_ usually extend the idea has very little relevance
here. And never mind the fact that you've decided below that
you're wrong. A much more important fact is this:
Even if it _were_ true that the notion of convexity is often
extended in the way you say you usually extend it, that
still would not show that Robert was wrong (!!!).
Because when he said what he said he was referring
to the _previous_ non-extended notion.
David C. Ullrich
<ma...@rdrop.remove.com> wrote:
>On Wed, 9 Mar 2011, David C. Ullrich wrote:
>> <ma...@rdrop.remove.com> wrote:
>>>>
>>>>>>> If U is an open set of a topological vector space, is the
>>>>>>> convex hull of U open?
>>>>>>
>>>>>> Lemma. If A is open then A + B = {a + b : a in A, b in B}
>>>>>> is open.
>>>>>>
>>>> It's still immediate from the Lemma that the convex
>>>> hull of an open set is open.
>
>It's immediate from the fact that the
>interior of a convex set is convex.
_What_ is immediate from the fact that the interior of
a convex set is convex?
Are you saying that (i) the fact that the convex hull of
an open set is open is immediate from (ii) the fact that
the interior of a convex set is convex?
Ok, yes, (i) is immediate from (ii). Took me a second, but yes.
_But_ before saying that this gives a simpler proof of (i)
than the proof I gave, avoiding those nonexistent
"loose ends", you need to explain exactly how you prove
(ii). I don't see how one proves (ii) except by an argument
of essentially identical complexity to the proof of (i)
that I gave.
How _do_ you prove (ii)? (I'm not asking for help with
the proof, I want to see the proof of (ii) that _you_
have in mind, that's so much simpler than the simple
proof of (i) that I gave.)
>
>>>> Say O is open and x is in the convex hull of O. There exist x_1, ...
>>>> x_n in O and t_j >= 0 with sum t_j = 1 and sum t_j x_j = x.
>
>Is that true of O if O is any subset or only if O is open?
>
>>>> Not all the t_j vanish. Say t_1 <> 0. Let
>>>> S = t_1 O + sum_{j>1} t_j x_j.
>>>>
>>>> The lemma shows that S is open, and it's clear that
>>>> x is in S and S is a subset of the convex hull of O.
>>>>
>>> It's not clear that S subset convex hull O.
>>
>> Huh??? What do you think the definition of the convex hull
>> of O is?
>>
>The smallest convex set containing O.
Right. And it's trivial to show from that definition that the
convex hull of O is equal to the set of all sums
sum_{j=1}^n t_j x_j where t_j >= 0, sum t_j = 1,
and x_j is in O.
(Hint: Let S be the set of all those sums. Show that
S is convex, and that every convex set containing
O also contains S. The second is literally trivial, while
the first is very easy.)
>
>>>>>>> If U is a convex set, is the closure of U convex?
>>>>>> Yes, this is trivial.
>>>>
>>>> As often happens, things are most obvious if we think
>>>> in terms of nets. Say C is convex, K is the closure of C,
>>>> x , y are in K and 0 < t < 1.
>>>>
>>>> There exist nets x_j and y_j in C with x_j -> x and
>>>> y_j -> y. Continuity of the vector-space operations
>>>> shows that t x_j + (1-t) y_j -> tx + (1-t)y; hence
>>>> tx + (1-t)y is in K.
>>>>
>>>> (Hmm, can you spot the slight technicality I'm
>>>> overlooking there and fix it?)
>>>
>>> For all y, x_j + (1 - t)y_j in C.
>>
>> No, that's not a technicality I was overlooking,
>> that's something that's so utterly obvious that
>> it didn't need to be stated.
>
>You've assume that the two nets have the same index.
Right. There are two ways to fix that. One: There's
a simple general construction, involving an appropriate
order on the product of the two index sets, that
shows that in a situation like this we can obtain
nets that do have the same index set.
Second, in the context of topological vector spaces,
the natural two index sets _are_ the same.
Or canonically isomorphic, anyway: If x is in the
closure of S then there is a net x_j in S converging
to x. The natural index set here is the set of
all neighborhoods of x, ordered by reverse inclusion.
Now in a topological vector space a neighborhood of
x is precisely x + A, where A is a neighbrhod of 0.
And similarly for y...
>
>
>
>Let M be module with an ordered ring.
>For example, R^2 over the integers.
>What definition is used for a convex set of M?
What makes you think there _is_ a standard definition?
Rob Johnson
I know that all three of these statement are true, but its seemed
as if you were saying that the notion of a basis for a vector space
was no different than that for a Hilbert space, but then you said
>This depends on what you mean by "basis". There are
>various definitions of the word in various contexts,
>and at least two of them apply here.
So it does seem "that the way a basis in a Hilbbert space works is
not the same as the way a basis in other vector spaces work."
These three questions were posed simply to show this difference.
>>2. f(x) is in the Hilbert space L^2(R/Z).
>>
>>3. f(x) cannot be written as a finite linear combination of
>> elements of U.
>>
>>A Hilbert space is a complete inner product space, which is a vector
>>space with extra structure that allows limits. It is not a vector
>>space,
>
>A Hilbert space is not a vector space?
>
>Huh???????????????????????????????????????
Okay, I overstepped a bit there. The definition only requires that
a vector space be an abelian group under addition and a module over
its field. It doesn't make any requirements about a basis or how a
basis behaves. So, yes, a Hilbert space is a vector space.
What I was referring to was the way that a basis works in each. In
a vector space, all elements are finite linear combinations of basis
elements. In a Hilbert space, elements are possibly infinite linear
combinations of basis elements. U is a basis of L^2(R/Z) as a
Hilbert space, but not as a vector space. As a vector space,
L^2(R/Z) has a Hamel basis, but it is not U.
Yes, as I said in my previous message, I now realize that sigma-
convexity is different than (finite) convexity. My error was thinking
that Robert was talking about sigma-convexity.
So you don't need to convince me that Robert's counter-example works.
Now that I know he is not talking about sigma-convexity, I know that
it works.
Rob Johnson <r...@trash.whim.org>
William Elliot
> <ma...@rdrop.remove.com> wrote:
>
>>>>>>>> If U is an open set of a topological vector space, is the
>>>>>>>> convex hull of U open?
>>>>>>>
>>>>>>> Lemma. If A is open then A + B = {a + b : a in A, b in B}
>>>>>>> is open.
>
> an open set is open is immediate from (ii) the fact that
> the interior of a convex set is convex?
>
> Ok, yes, (i) is immediate from (ii). Took me a second, but yes.
>
> _But_ before saying that this gives a simpler proof of (i)
> than the proof I gave, avoiding those nonexistent
> "loose ends", you need to explain exactly how you prove
> (ii). I don't see how one proves (ii) except by an argument
> of essentially identical complexity to the proof of (i)
> that I gave.
>
> How _do_ you prove (ii)? (I'm not asking for help with
> the proof, I want to see the proof of (ii) that _you_
> have in mind, that's so much simpler than the simple
> proof of (i) that I gave.)
It's packed in the bottom of the post.
>>>>> Say O is open and x is in the convex hull of O. There exist x_1, ...
>>>>> x_n in O and t_j >= 0 with sum t_j = 1 and sum t_j x_j = x.
>>
>> Is that true of O if O is any subset or only if O is open?
>>
>>>>> Not all the t_j vanish. Say t_1 <> 0. Let
>>>>> S = t_1 O + sum_{j>1} t_j x_j.
>>>>>
>>>>> The lemma shows that S is open, and it's clear that
>>>>> x is in S and S is a subset of the convex hull of O.
>>>>>
>>>> It's not clear that S subset convex hull O.
>>>
>>> Huh??? What do you think the definition of the convex hull
>>> of O is?
>>>
>> The smallest convex set containing O.
>
> Right. And it's trivial to show from that definition that the
> convex hull of O is equal to the set of all sums
> sum_{j=1}^n t_j x_j where t_j >= 0, sum t_j = 1,
> and x_j is in O.
This is an interesting construction.
Is n fixed for the TVS or variable?
> (Hint: Let S be the set of all those sums. Show that
> S is convex, and that every convex set containing
> O also contains S. The second is literally trivial, while
> the first is very easy.)
>>>>>>>> If U is a convex set, is the closure of U convex?
>>>>>>> Yes, this is trivial.
>>>>>
>>>>> As often happens, things are most obvious if we think
>>>>> in terms of nets. Say C is convex, K is the closure of C,
>>>>> x , y are in K and 0 < t < 1.
>>>>>
>>>>> There exist nets x_j and y_j in C with x_j -> x and
>>>>> y_j -> y. Continuity of the vector-space operations
>>>>> shows that t x_j + (1-t) y_j -> tx + (1-t)y; hence
>>>>> tx + (1-t)y is in K.
>>>>>
>>>>> (Hmm, can you spot the slight technicality I'm
>>>>> overlooking there and fix it?)
>> You've assume that the two nets have the same index.
>
> Right. There are two ways to fix that. One: There's
> a simple general construction, involving an appropriate
> order on the product of the two index sets, that
> shows that in a situation like this we can obtain
> nets that do have the same index set.
>
Not so bad as compared to subnets.
> Second, in the context of topological vector spaces,
> the natural two index sets _are_ the same.
> Or canonically isomorphic, anyway: If x is in the
> closure of S then there is a net x_j in S converging
> to x. The natural index set here is the set of
> all neighborhoods of x, ordered by reverse inclusion.
>
> Now in a topological vector space a neighborhood of
> x is precisely x + A, where A is a neighbrhod of 0.
> And similarly for y...
Yes, topological groups are homogenous by the homeomorphism
f:G -> G, x -> a + x.
David C. Ullrich
<r...@trash.whim.org> wrote:
I don't see how you get there from what I said.
I think the problem is that you're taking my statements about
the meanings of words and interpreting them as statments
about mathematical facts. There's a big difference.
>These three questions were posed simply to show this difference.
>
>>>2. f(x) is in the Hilbert space L^2(R/Z).
>>>
>>>3. f(x) cannot be written as a finite linear combination of
>>> elements of U.
>>>
>>>A Hilbert space is a complete inner product space, which is a vector
>>>space with extra structure that allows limits. It is not a vector
>>>space,
>>
>>A Hilbert space is not a vector space?
>>
>>Huh???????????????????????????????????????
>
>Okay, I overstepped a bit there.
Relieved to see you say that.
> The definition only requires that
>a vector space be an abelian group under addition and a module over
>its field. It doesn't make any requirements about a basis or how a
>basis behaves. So, yes, a Hilbert space is a vector space.
>
>What I was referring to was the way that a basis works in each. In
>a vector space, all elements are finite linear combinations of basis
>elements. In a Hilbert space, elements are possibly infinite linear
>combinations of basis elements. U is a basis of L^2(R/Z) as a
>Hilbert space, but not as a vector space. As a vector space,
>L^2(R/Z) has a Hamel basis, but it is not U.
This is not a difference in the way a basis works. It is a difference
in what the word "basis" (typically) means. It's not clear to me
whether you see what I mean by this - here's a silly example:
Imagine that the English language has just been revised, so
that the word "apple" refers to the animals currently known
as cats - that's the only change. Then both the following
statments are true:
(i) The sentence "An apple is a fruit" is false.
(ii) An apple is a fruit.
How can they both be true? Both (i) and (ii) are written
in old-English, as all my posts are. But the sentence
quoted in (i) is a new-English sentence. Changing
the meaning of the word "apple" has no effect on
the _fact_ that an apple is a fruit, it just changes
the sequence of words one uses to express that fact.
David C. Ullrich
<ma...@rdrop.remove.com> wrote:
>On Thu, 10 Mar 2011, David C. Ullrich wrote:
>> <ma...@rdrop.remove.com> wrote:
>>
>>>>>>>>> If U is an open set of a topological vector space, is the
>>>>>>>>> convex hull of U open?
>>>>>>>>
>>>>>>>> Lemma. If A is open then A + B = {a + b : a in A, b in B}
>>>>>>>> is open.
>>
>> Are you saying that (i) the fact that the convex hull of
>> an open set is open is immediate from (ii) the fact that
>> the interior of a convex set is convex?
>>
>> Ok, yes, (i) is immediate from (ii). Took me a second, but yes.
>>
>> _But_ before saying that this gives a simpler proof of (i)
>> than the proof I gave, avoiding those nonexistent
>> "loose ends", you need to explain exactly how you prove
>> (ii). I don't see how one proves (ii) except by an argument
>> of essentially identical complexity to the proof of (i)
>> that I gave.
>>
>It's done and finished. Yours still requires a proof of a lemma.
What lemma?
>> How _do_ you prove (ii)? (I'm not asking for help with
>> the proof, I want to see the proof of (ii) that _you_
>> have in mind, that's so much simpler than the simple
>> proof of (i) that I gave.)
>
>It's packed in the bottom of the post.
And that "packed" proof is supposed to be simpler
somehow? Curious notion of "simple" - the
incoherent non-English "packed" version is as long
as the proof I gave.
Why don't you write out a version in English so we
can actually see which one is simpler?
>>>>>> Say O is open and x is in the convex hull of O. There exist x_1, ...
>>>>>> x_n in O and t_j >= 0 with sum t_j = 1 and sum t_j x_j = x.
>>>
>>> Is that true of O if O is any subset or only if O is open?
>>>
>>>>>> Not all the t_j vanish. Say t_1 <> 0. Let
>>>>>> S = t_1 O + sum_{j>1} t_j x_j.
>>>>>>
>>>>>> The lemma shows that S is open, and it's clear that
>>>>>> x is in S and S is a subset of the convex hull of O.
>>>>>>
>>>>> It's not clear that S subset convex hull O.
>>>>
>>>> Huh??? What do you think the definition of the convex hull
>>>> of O is?
>>>>
>>> The smallest convex set containing O.
>>
>> Right. And it's trivial to show from that definition that the
>> convex hull of O is equal to the set of all sums
>> sum_{j=1}^n t_j x_j where t_j >= 0, sum t_j = 1,
>> and x_j is in O.
>
>This is an interesting construction.
>Is n fixed for the TVS
Of course not.
It's remarkable how you can "correct" people on topics
that you don't know the first thing about. Yes, the
fact that those two definitions of convex hull are the
same is the very first thing - it's not an interesting
construction, it's an absolute triviality.
Rob Johnson
I am not quite sure what you mean by "how a basis works". I know
what a basis means:
A basis for a vector space is a minimal set of elements of that
vector space so that any element in that vector space is a finite
linear combination of those basis elements.
A basis for a Hilbert space is a minimal set of elements of that
Hilbert space so that any element of that Hilbert space is a
convergent linear combination of those basis elements.
These definitions are similar, only exchanging the words "finite"
and "convergent".
Bases work similarly in that we use linear combinations of basis
elements in both settings. However, they work differently depending
on whether or not we have enough topological structure to allow
limits so that we can add infinitely many basis elements.
The definitions are similar but different, and how bases work is
also similar but different.
Perhaps you had something different in mind.
>Imagine that the English language has just been revised, so
>that the word "apple" refers to the animals currently known
>as cats - that's the only change. Then both the following
>statments are true:
>
>(i) The sentence "An apple is a fruit" is false.
>
>(ii) An apple is a fruit.
>
>How can they both be true? Both (i) and (ii) are written
>in old-English, as all my posts are. But the sentence
>quoted in (i) is a new-English sentence. Changing
>the meaning of the word "apple" has no effect on
>the _fact_ that an apple is a fruit, it just changes
>the sequence of words one uses to express that fact.
My head hurts, and all my apples are furry.
William Elliot
> <ma...@rdrop.remove.com> wrote:
>>>
>>>>>>>>>> If U is an open set of a topological vector space, is the
>>>>>>>>>> convex hull of U open?
>>>>>>>>>
>>>>>>>>> Lemma. If A is open then A + B = {a + b : a in A, b in B} is
>>>>>>>>> open.
>>>
>>>>>>> ... x_n in O and t_j >= 0 with sum t_j = 1 and sum t_j x_j = x.
>>>>
>>>>>>> S = t_1 O + sum_{j>1} t_j x_j.
>>>>>>>
>>>>>>> The lemma shows that S is open, and it's clear that
>>>>>>> x is in S and S is a subset of the convex hull of O.
>>>>>>>
>>>>>> It's not clear that S subset convex hull O.
>>>>>
>>>>> Huh??? What do you think the definition of the convex hull
>>>>> of O is?
>>>>>
>>>> The smallest convex set containing O.
>>>
>>> Right. And it's trivial to show from that definition that the
>>> convex hull of O is equal to the set of all sums
>>> sum_{j=1}^n t_j x_j where t_j >= 0, sum t_j = 1,
>>> and x_j is in O.
>>
>> This is an interesting construction.
>> Is n fixed for the TVS
> Of course not.
>
If A is any subset of a vector space (F,S), then
x in convex hull A iff
some j in N, x1,.. xj in A, t1,.. tj >= 0
with t1 +..+ tj = 1, x = t1.x1 +..+ tj.xj.
Why isn't F = R, S = R^R and A = { f_r | r in R } \/ { f0 }
where f0 is the zero funciton, f_r(x) = 0 if x /= r
and f_r(r) = 1, a counter example?
Let H = { x | some j in N, x1,.. xj in A, t1,.. tj >= 0
with t1 +..+ tj = 1, x = t1.x1 +..+ tj.xj }.
To show that H is the convex hull of A, one needs to show
A subset H, H is convex and H subset convex hull A.
A subset H is trivial. To show H is convex, assume
x = sum_j tj.xj, y = sum_k sk.yk in H, r in [0,1].
rx + (1 - r)y = sum_j rtj.xj + sum_k (1 - r)sk.yk in H
because
sum_j rtj + sum_k (1 - r)sk = r + 1 - r = 1.
Finally what are the trivial steps needed to show
H subset convex hull A?
BTW, how is "convex hull A" printed in text books?
Let (xj)_j and (yk)_k be two nets with domains J and K, resp.
Define a net z over I = JxK by z(j,k) = (xj,yk)
where JxK is coordinate wised ordered.
The nets p1 o z and p2 o z are two nets with the same domain.
In the event that xj -> x and yk -> y,
z_i -> (x,y), p1.z_i -> x, p2.z_i -> y.
In addition, p1.z_i + p2.z_i -> x + y which can loosely be stated as
x_i + y_i -> x + y.
Classified ppf. Inquire of FOI for illucidation.
David C. Ullrich
<r...@trash.whim.org> wrote:
Yes, I had something different in mind! I was talking about
the pure-vector-space notion of "basis". The definition of
_that_ notion is the same in a Hilbert space as in any other
vector space.
>
>>Imagine that the English language has just been revised, so
>>that the word "apple" refers to the animals currently known
>>as cats - that's the only change. Then both the following
>>statments are true:
>>
>>(i) The sentence "An apple is a fruit" is false.
>>
>>(ii) An apple is a fruit.
>>
>>How can they both be true? Both (i) and (ii) are written
>>in old-English, as all my posts are. But the sentence
>>quoted in (i) is a new-English sentence. Changing
>>the meaning of the word "apple" has no effect on
>>the _fact_ that an apple is a fruit, it just changes
>>the sequence of words one uses to express that fact.
>
>My head hurts, and all my apples are furry.
Maybe you'd prefer the old joke:
Q. How many legs does a horse have if you call the tail a leg?
A. Four. Calling the tail a leg doesn't make it a leg.
David C. Ullrich
That's not "the problem", but yes, that's the extremely
obvious first thing one notes about convex hulls.
>Why isn't F = R, S = R^R and A = { f_r | r in R } \/ { f0 }
>where f0 is the zero funciton, f_r(x) = 0 if x /= r
>and f_r(r) = 1, a counter example?
How can I possibly answer that? It's _not_ a counterexample,
since there are none. Whatever reason you have for thinking
it _is_ a counterexample is wrong, but I can't possibly say
what your error is unless you explain to me why you think
it _is_ a counterexample...
>Let H = { x | some j in N, x1,.. xj in A, t1,.. tj >= 0
> with t1 +..+ tj = 1, x = t1.x1 +..+ tj.xj }.
>
>To show that H is the convex hull of A, one needs to show
>A subset H, H is convex and H subset convex hull A.
>
>A subset H is trivial. To show H is convex, assume
>x = sum_j tj.xj, y = sum_k sk.yk in H, r in [0,1].
>
>rx + (1 - r)y = sum_j rtj.xj + sum_k (1 - r)sk.yk in H
>because
>sum_j rtj + sum_k (1 - r)sk = r + 1 - r = 1.
>
>Finally what are the trivial steps needed to show
> H subset convex hull A?
Formally, it's a trivial induction on n. Informally:
Say t_1 + t_2 + t_3 = 1, t_j >= 0, x_j in A.
If t_3 = 1 then t_1 = t_2 = 0, so
t_1 x_1 + t_2 x_2 + t_3 x_3 = x_3
is in the convex hull of A.
Suppose that t_3 > 0. Let s_1 = t_1 / ( 1 - t_3),
s_2 = t_2 /(1 - t_3). Then s_1 + s_2 = 1, so
s_1 x_1 + s_2 x_2 is in the convex hull of A.
But
t_1 x_1 + t_2 x_2 + t_3 x_3
= (1 - t_3) (s_1 x_1 + s_2 x_2) + t_3 x_3,
so s_1 x_1 + s_2 x_2 in the convex hull of A
and x_3 in the convex hul of A imply that
t_1 x_1 + t_2 x_2 + t_3 x_3 is in the convex
hull of A.
>BTW, how is "convex hull A" printed in text books?
That varies from author to author.
Rob Johnson
I must be misunderstanding something here.
In the "pure-vector-space" notion of a basis, each vector must be a
finite linear combination of the basis elements. Such a basis, a
Hamel basis, exists for separable Hilbert spaces, such as L^2(R/Z).
However, I don't think Robert was talking about a "pure-vector-space"
since he wanted the "orthonormal basis {u_1,u_2,...}" to be closed,
and he uses the inner product from the Hilbert space to show that it
is. Furthermore, the orthonormal basis {u_1,u_2,...} is not a Hamel
basis for L^2(R/Z).
>>>Imagine that the English language has just been revised, so
>>>that the word "apple" refers to the animals currently known
>>>as cats - that's the only change. Then both the following
>>>statments are true:
>>>
>>>(i) The sentence "An apple is a fruit" is false.
>>>
>>>(ii) An apple is a fruit.
>>>
>>>How can they both be true? Both (i) and (ii) are written
>>>in old-English, as all my posts are. But the sentence
>>>quoted in (i) is a new-English sentence. Changing
>>>the meaning of the word "apple" has no effect on
>>>the _fact_ that an apple is a fruit, it just changes
>>>the sequence of words one uses to express that fact.
>>
>>My head hurts, and all my apples are furry.
>
>Maybe you'd prefer the old joke:
>
>Q. How many legs does a horse have if you call the tail a leg?
>
>A. Four. Calling the tail a leg doesn't make it a leg.
Is there something that I am calling by the wrong name? Your last
two stories/jokes seem to imply so. I realize that sigma-convex is
not the same as convex. I am not suggesting otherwise.
William Elliot
> <ma...@rdrop.remove.com> wrote:
>
>>>>>>>>>>>> If U is an open set of a topological vector space, is the
>>>>>>>>>>>> convex hull of U open?
>>>>>>>>>>>
>>>>>>>>>>> Lemma. If A is open then A + B = {a + b : a in A, b in B} is
>>>>>>>>>>> open.
>>>>>
>>>>>>>>> Say O is open and x is in the convex hull of O. There exist x_1,
>>>>>>>>> ... x_n in O and t_j >= 0 with sum t_j = 1 and sum t_j x_j = x.
>>>>>>
>>>>>>>>> Not all the t_j vanish. Say t_1 <> 0. Let
>>>>>>>>> S = t_1 O + sum_{j>1} t_j x_j.
>>>>>>>>>
>>>>>>>>> The lemma shows that S is open, and it's clear that x is in S
>>>>>>>>> and S is a subset of the convex hull of O.
>>>>>>>>>
>>>>>>>> It's not clear that S subset convex hull O.
>>>>>> The smallest convex set containing O.
>>>>>
>>>>> Right. And it's trivial to show from that definition that the
>>>>> convex hull of O is equal to the set of all sums
>>>>> sum_{j=1}^n t_j x_j where t_j >= 0, sum t_j = 1,
>>>>> and x_j is in O.
>> This then is the problem?
>>
>> If A is any subset of a vector space (F,S), then
>> x in convex hull A iff
>> some j in N, x1,.. xj in A, t1,.. tj >= 0
>> with t1 +..+ tj = 1, x = t1.x1 +..+ tj.xj.
>
> That's not "the problem", but yes, that's the extremely
> obvious first thing one notes about convex hulls.
>
Ok, I get it - accounting tedium. Anyway, what's the merit of
the proposition we're discussing other that to show the convex hull
of an open set is open. Are there other theorems that use that
construction for the convex hull?
>> BTW, how is "convex hull A" printed in text books?
> That varies from author to author.
What are a couple that you like?
Consider R^R over the reals and let A = { f | f(R) subset {0,1} }.
A is the set of corners of a unit cube. What's convex hull A?
Shouldn't it be C = { f | f(R) subset [0,1] }?
Is convex hull A actually a proper subset of C?
If so, is convex hull A a dense subset of C?
>> Let (xj)_j and (yk)_k be two nets with domains J and K, resp.
>>
>> Define a net z over I = JxK by z(j,k) = (xj,yk)
>> where JxK is coordinate wised ordered.
>>
>> The nets p1 o z and p2 o z are two nets with the same domain.
>>
>> In the event that xj -> x and yk -> y,
>> z_i -> (x,y), p1.z_i -> x, p2.z_i -> y.
>>
>> In addition, p1.z_i + p2.z_i -> x + y which can loosely be stated as
>> x_i + y_i -> x + y.
>>
quasi
<ma...@rdrop.remove.com> wrote:
>Consider R^R over the reals and let
> A = { f | f(R) subset {0,1} }.
>A is the set of corners of a unit cube.
>What's convex hull A?
>Shouldn't it be C = { f | f(R) subset [0,1] }?
No.
>Is convex hull A actually a proper subset of C?
Yes, it's a proper subset.
The convex hull of A is the subset of R^R consisting of all
functions f in R^R such that f(R) is a _finite_ subset of
the closed interval [0,1].
>If so, is convex hull A a dense subset of C?
What's your assumed topology on R^R?
If it's the product topology, then yes, I think it's dense.
The open subsets of C (with the relative topology inherited
from the product topology on R^R) are so big that none of
them can avoid having elements of the convex hull of A.
quasi
William Elliot
for the proposition that was under immediate discussion.
David C. Ullrich
<r...@trash.whim.org> wrote:
I can't think of a way to explain the point I'm
trying to make here any more clearly than the
last two times.
David C. Ullrich
<ma...@rdrop.remove.com> wrote:
Erm, you really don't think that knowing exactly what the elements of
the convex hull of a set _are_ could be useful, unless I come
up with a specific example?
I mean there's a question below the answer to which would
have been obvious, if you'd thought for a second about
the fact that the convex hull of A is the set of all
"convex combinations" of the elements of A.
>>> BTW, how is "convex hull A" printed in text books?
>> That varies from author to author.
>
>What are a couple that you like?
>
>Consider R^R over the reals and let A = { f | f(R) subset {0,1} }.
>A is the set of corners of a unit cube. What's convex hull A?
>Shouldn't it be C = { f | f(R) subset [0,1] }?
Don't know about "should be", but it's not that.
It's obviously not that, if you know the first
thing about what the convex hull of a set is.
(Hint: I've explained above, a few times, what
that "first thing" is.)
>Is convex hull A actually a proper subset of C?
Yes.
>If so, is convex hull A a dense subset of C?
Yes, if you're talking about the product topology.
quasi
>On Sun, 13 Mar 2011 20:25:03 -0700, William Elliot
><ma...@rdrop.remove.com> wrote:
>
>>Consider R^R over the reals and let
>> A = { f | f(R) subset {0,1} }.
>>A is the set of corners of a unit cube.
>>What's convex hull A?
>>Shouldn't it be C = { f | f(R) subset [0,1] }?
>
>No.
>
>>Is convex hull A actually a proper subset of C?
>
>Yes, it's a proper subset.
>
>The convex hull of A is the subset of R^R consisting of all
>functions f in R^R such that f(R) is a _finite_ subset of
>the closed interval [0,1].
Correction:
The convex hull of A is the subset of R^R consisting of all
functions f in R^R such that f(R) is a _finite_ subset of
the closed interval [0,1] and such that the sum of the
range values is 1.
>>If so, is convex hull A a dense subset of C?
>
>What's your assumed topology on R^R?
>
>If it's the product topology, then yes, I think it's dense.
>The open subsets of C (with the relative topology inherited
>from the product topology on R^R) are so big that none of
>them can avoid having elements of the convex hull of A.
Correction:
If it's the product topology, then no, it's not dense.
If 1/2 < q <= 1, the constant function f:R->R with f(x) = q for
all x in R has an open neighborhood (containing q) disjoint
from the convex hull of A.
quasi
quasi
>On Mon, 14 Mar 2011 02:59:31 -0500, quasi <qu...@null.set> wrote:
>
>>On Sun, 13 Mar 2011 20:25:03 -0700, William Elliot
>><ma...@rdrop.remove.com> wrote:
>>
>>>Consider R^R over the reals and let
>>> A = { f | f(R) subset {0,1} }.
>>>A is the set of corners of a unit cube.
>>>What's convex hull A?
>>>Shouldn't it be C = { f | f(R) subset [0,1] }?
>>
>>No.
>>
>>>Is convex hull A actually a proper subset of C?
>>
>>Yes, it's a proper subset.
>>
>>The convex hull of A is the subset of R^R consisting of all
>>functions f in R^R such that f(R) is a _finite_ subset of
>>the closed interval [0,1].
>
>Correction:
>
>The convex hull of A is the subset of R^R consisting of all
>functions f in R^R such that f(R) is a _finite_ subset of
>the closed interval [0,1] and such that the sum of the
>range values is 1.
2nd correction:
The convex hull of A is the set of all function f:R->R
such that f(R) is a finite subset of the closed interval
[0,1] and such that the sum of the distinct range values
is at most 1.
>>>If so, is convex hull A a dense subset of C?
>>
>>What's your assumed topology on R^R?
>>
>>If it's the product topology, then yes, I think it's dense.
>>The open subsets of C (with the relative topology inherited
>>from the product topology on R^R) are so big that none of
>>them can avoid having elements of the convex hull of A.
>
>Correction:
>
>If it's the product topology, then no, it's not dense.
>
>If 1/2 < q <= 1, the constant function f:R->R with f(x) = q for
>all x in R has an open neighborhood (containing q) disjoint
>from the convex hull of A.
2nd correction:
If it's the product topology, then no, it's not dense.
Let f:R->R be any function such that f(R) is a finite subset
of the closed interval [0,1] and such that the sum of the
distinct range values of f is greater than 1. Then there is
an open subset of containing f but disjoint from the convex
hull of A. It follows that the convex hull of A is not dense
in C.
I think it's right now.
quasi
quasi
In fact, let U be the complement in C of the convex hull of A.
Then U is the set of f in C such that there exists a finite
subset of f(R) with sum greater than 1. Clearly U is nonempty.
Moreover U is open (and the convex hull of A is closed).
Note that for f to be in U, f(R) need not be finite -- just so
long as the sum of some finite subset of f(R) exceeds 1. In
particular, if f(R) is uncountable, then it's automatically
true that f is in U.
quasi
quasi
Correction:
Let H be the convex hull of A.
While it is true that C\H has nonempty interior, as far as I
can see, C\H is not open, and thus, the H is not closed.
Let B be the set of f in C such that there exists a finite
subset of f(R) with sum greater than 1. Then B is open and
nonempty, and the interior of C\H is equal to B, but B is a
proper subset of C\H.
>Note that for f to be in U, f(R) need not be finite -- just so
>long as the sum of some finite subset of f(R) exceeds 1. In
>particular, if f(R) is uncountable, then it's automatically
>true that f is in U.
In the lines above, replace U by V.
Remarks:
I'm not so sure now that I have it right -- maybe someone can
check these claims. I'll summarize them ...
Let X = R^R, with the product topology, regarded as a topological
vector space over R.
Let C = {f in X | f(R) is a subset of [0,1]}.
Let A = {f in C | f(R) is a subset of {0,1}}.
Let H be the convex hull of A.
Let B = {f in C | some finite subset of f(R) has sum > 1}
Claims:
(1) H is equal to the set of f in C such that f(R) is finite
and the sum of the distinct values in f(R) is at most 1.
(2) B is a nonempty open subset of C which is disjoint from H
(hence H is not dense in C).
(3) H is not closed.
(4) The interior of C\H is equal to B.
Yes?
quasi
David C. Ullrich
>[...]
>
>I'm not so sure now that I have it right -- maybe someone can
>check these claims. I'll summarize them ...
>
>Let X = R^R, with the product topology, regarded as a topological
>vector space over R.
>
>Let C = {f in X | f(R) is a subset of [0,1]}.
>
>Let A = {f in C | f(R) is a subset of {0,1}}.
>
>Let H be the convex hull of A.
>
>Let B = {f in C | some finite subset of f(R) has sum > 1}
>
>Claims:
>
>(1) H is equal to the set of f in C such that f(R) is finite
>and the sum of the distinct values in f(R) is at most 1.
No, H is the set of f in C such that f(R) is finite, period.
Hint: Given 0 = v_0 < ... < v_n = 1, let t_j = v_j - v_{j-1}.
>(2) B is a nonempty open subset of C which is disjoint from H
>(hence H is not dense in C).
>
>(3) H is not closed.
True. In fact the closure of H is exactly C.
Hint: Suppose f is in C and N is a neighborhood
of f. By definition of the product topology there
exists a _finite_ set F such that if |g(x) - f(x)| < eps
for all x in F then g is in N. You can certainly find
such g in H (once you get straight what H is).
quasi
<ull...@math.okstate.edu> wrote:
>On Tue, 15 Mar 2011 06:31:03 -0500, quasi <qu...@null.set> wrote:
>
>>[...]
>>
>>I'm not so sure now that I have it right -- maybe someone can
>>check these claims. I'll summarize them ...
>>
>>Let X = R^R, with the product topology, regarded as a topological
>>vector space over R.
>>
>>Let C = {f in X | f(R) is a subset of [0,1]}.
>>
>>Let A = {f in C | f(R) is a subset of {0,1}}.
>>
>>Let H be the convex hull of A.
>>
>>Let B = {f in C | some finite subset of f(R) has sum > 1}
>>
>>Claims:
>>
>>(1) H is equal to the set of f in C such that f(R) is finite
>>and the sum of the distinct values in f(R) is at most 1.
>
>No, H is the set of f in C such that f(R) is finite, period.
That's what I said originally (in my first reply).
Somehow my visualization of the potential convex combinations
changed -- why, I don't know.
>Hint: Given 0 = v_0 < ... < v_n = 1, let t_j = v_j - v_{j-1}.
Yes, I see it now.
Just looking at a convex combination of 2 elements dispels
the illusion.
Let f1,f2 be in A, and let
f = t1*f1 + t2*f2
where t1,t2 are nonnegative reals with t1 + t2 = 1.
Then f(R) is a nonempty subset of {0,t1,t2,1} but with
appropriate choices of f1,f2, any such subsets is possible.
>>(2) B is a nonempty open subset of C which is disjoint from H
>>(hence H is not dense in C).
>>
>>(3) H is not closed.
>
>True. In fact the closure of H is exactly C.
Right.
>Hint: Suppose f is in C and N is a neighborhood
>of f. By definition of the product topology there
>exists a _finite_ set F such that if |g(x) - f(x)| < eps
>for all x in F then g is in N. You can certainly find
>such g in H (once you get straight what H is).
Yes, that's exactly the way I conceptualized it in my
first reply.
Thanks.
quasi