-- The "most natural" completion of a metric space
omega
Met := (Objects = metric spaces, Morphisms = isometries);
CMet := (Objects = complete metric spaces, Morphisms = isometries)
Is there a covariant functor F : Met --> CMet satisfying :
1) for each metric space M there is an isometry j_M : M --> F(M) with
dense range, and
2) If T is complete, then F(T) = T ?
Thx in advance.
P.S. [not necessarily related] BTW, are Met and CMet equivalent categories ?
Musatov
[PDF]
A Study of Global Inference Algorithms in Multi-Document Summarization
File Format: PDF/Adobe Acrobat - View as HTML
defining a general framework and proving that inference in it is NP-
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ab
the completion of any metric space is unique up to isometry, so looking for a "most natural" completion is not necessary
omega
> forgive me if what i'm saying is trivial to you, but
>
> the completion of any metric space is unique up to
> isometry,
Very true. And this is the "completion functor" between the skeletons [Met] and [CMet], i.e. when
any two metric spaces are identified if they are isometric.
> so looking for a "most natural" completion
> is not necessary
My point is the completion of a given metric space is in fact a class of isometric but different
"concrete relizations", and the problem is : how to *choose* among them still in a *functorial* way,
without using any nontrivial equivalence.
E.g., if N is a normed space, one can view its completion either as the closure of N into N**, or
as the closure of N into C(S), where S is the closed unit ball of N*, endowed with the w*-topology.
Or in many other (equivalent, but distinct) ways (e.g., using the construction with Cauchy sequences).
Which is "the most adequate" to be nominated as *the* completion of N ?
So that, I'm looking for a completion functor that preserves the "concrete nature of the objects", i.e.,
not identifying two different metric spaces, even that they are isometric. And also preserving the
complete metric spaces. (I do not know if such a "lifting" is possible.)
David C. Ullrich
I'm really not sure what the requirement is, but have you condsidered
the standard construction of the completion of a metric space?
Say X is a metric space. Let C be the space of Cauchy sequences of
elements of X. Say two sequences (x_n), (y_n) in C are equivalent
if d(x_n, y_n) -> 0. Then the quotient of C by this equivalence
relation is the completion of X (with the metric defined in the
obvious way and with the obvious embedding of X; both
those are left as exercises if they're not obvious...)
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
A N Niel
Ullrich <dull...@sprynet.com> wrote:
> I'm really not sure what the requirement is, but have you condsidered
> the standard construction of the completion of a metric space?
With his requirements it looks like what he has to do is this ( done
with the example of the rationals ) ... Take the usual completion of
the rationals, then remove the points that correspond to rationals and
replace them with the original rationals. Seems "less natural" to me.
Dave L. Renfro
> Let us define the following two concrete categories :
> Met := (Objects = metric spaces, Morphisms = isometries);
> CMet := (Objects = complete metric spaces, Morphisms = isometries)
>
> Is there a covariant functor F : Met --> CMet satisfying :
> 1) for each metric space M there is an isometry j_M : M --> F(M)
> with dense range, and 2) If T is complete, then F(T) = T ?
I've never known much category theory (at one time I was
familiar with the material in the chapter on categories in
Hungerford's "Algebra", but that was nearly 30 years ago
and I've had very little contact with category theory since
then), but I believe the completion of a metric space is
one of the constructions that Saunders MacLane gives as
an example (in "Categories for the Working Mathematician")
in his chapter on adjoint functors. I remember this because
I went through several advanced undergraduate math courses
in high school, in particular I had covered the completion
of a metric space (Robert Kasriel's "Undergraduate Topology")
and the quotient field of an integral domain (I. N. Herstein's
"Topics in Algebra"), and both constructions stuck me as
being very similar in many respects, so I asked some math
professors if there was any higher-level theory that these
two constructions were specific examples of, and no one
knew of any. Then, maybe three years later, I happened to
see both constructions mentioned in MacLane's "Categories
for the Working Mathematician", although by then my interests
had changed more to physics (and then, a few years later
still, I returned to math after I realized I had neither
the interest nor the ability to pursue mathematical physics),
and I've never looked back into this to see whether the
adjoint functor link creates at best a tenuous connection,
or something more, between extending the integers to the
rationals and extending the rationals to the reals.
Dave L. Renfro
omega
So that, consider for a metric space M the standard construction of its completion, say M',
together with the canonical embedding j : M --> M'.
Next, define F(M) := M \/ (M' \ j(M)) (disjoint union), endowed with a natural metric d.
[Namely, this metric preserves the original metric of M, the metric on M' \ j(M) inherited from
the constructed one on M', and d(m,m') = d_{M'}(j(m),m') if m in M and m' in M' \ j(M).]
It should works. Thank you.
> Seems "less natural" to me.
Why should be the above construction "less natural" than identifying an element q in Q with
the equivalence class of rational sequences converging to q (i.e., the equivalence class containing
the constant sequence (q, q, q, ...) ) ?
P.S. A weak point of your suggestion : one cannot have "multiplicativity", i.e.
F(M_1 x M_2) = F(M_1) x F(M_2), that is desirable.
KMF
This is a question from 'Analyst' in Topology
Atlas.
Mariano Suárez-Alvarez
Your question is then of a very uncategorical nature...
The "concrete nature of the objects" is something that
has no real menaing, so it is difficult to preserve it!
-- m
omega
> This is a question from 'Analyst' in Topology
> Atlas.
Actually, the "source" is the answer. My answer.
omega
> Your question is then of a very uncategorical
> nature...
> The "concrete nature of the objects" is something
> that
> has no real menaing, so it is difficult to preserve
> it!
>
> -- m
>
>
Then please ignore my further comments and restrict to
the original question. It is quite concrete, I think.
Namely, I'm looking for a "completion functor" F : Met --> CMet satisfying as many as possible desirable properties.
Mariano Suárez-Alvarez
There is exactly one sensible functor from
Met to CMet, the completion, and it is well-defined
up to isomorphism of functors. It preserves "as many
as possible desirable properties".
-- m
omega
Including among them, I hope :
1) UFj = U, and
2) UF(M_1 x M_2) = UF(M_1) x UF(M_2) for all M_1, M_2 in Ob(Met),
where U : CMet --> Set is the forgetful functor, j : CMet --> Met is the natural embedding,
and F : Met --> CMet is your sensible functor.
BTW, could you please tell me who is F(M), if M = (C[0,1], d), where "d" is the distance
generated by the L^2 -norm ? This should be a member of Ob(CMet), isn't it ?
Mariano Suárez-Alvarez
The functor is defined up to isomorphisms of functors, so
F(M) is quite flexible...
One way to construct the functor F is as follows.
If X is a metric space, let S(X) be the set of all
Cauchy sequences in X, and let ~ be the usual
equivalence relation on S(X), and let F(X) = S(X)/~
In other words, the usual construction of the completion
of X which is explained in every single relevant textbook
out there.
-- m
omega
"Sensible" and "quite flexible" are indeed very desirable properties. But not of a functor... ;-)
>
> One way to construct the functor F is as follows.
> If X is a metric space, let S(X) be the set of all
> Cauchy sequences in X, and let ~ be the usual
> equivalence relation on S(X), and let F(X) = S(X)/~
> In other words, the usual construction of the
> completion
> of X which is explained in every single relevant
> textbook
> out there.
And this triviality is satisfying 1) ?
>
> -- m
Mariano Suárez-Alvarez
Both of your conditions are not sensible conditions
to ask, for they just are motivated by your idea
of "preserving the nature of elements", which makes very
little sense.
-- m
Musatov
Musatov wrote:
I apologize for every off topic point, mean spirited comment, and
otherwise interruption.
Some things are true simply because of how beautiful they are. When
beauty presents itself it sometimes does so in efficiency. Sometimes
efficiency is the product of beautiful intricacy and function without
large amounts of effort. Like how ferns spiral out in a perfect
"Fibonacci" pattern or a snail shell holds it in perfect strength and
structural efficiency. The snail need not understand the mathematics
to benefit from the design and the plant need not awareness for each
throng of ferns to benefit from distributed sunlight.
To this I can only say there is a way in nature to discovering great
truths.
Here is an example:
http://m.flickr.com/photos/ninecats/2271191972/
Now on to the matter I seek out of the goodness of my heart to humbly
resolve for no benefit other than done to others.
There is something to the below text discernable to a studying mind
seeking to understand in a progressive stance.
Google Groups Alert for: Musatov proof
<http://www.docendi.org/proof-t249252.html> proof of the Kepler
Packing Problem is mostly a check on three ...
In the proof of KPP, we consider every Real Number as a length of a
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m (Musatov) ...
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-- The "most natural" completion of a metric space <http://
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Musatov marty.musa...@gmail.com sci math comp theory On Jul 5, 6:42
pm, ... inference in it is NP- hard. .... of exactly K. Proof: First
we observe that if ...
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David C. Ullrich
>> In article
>> <c9s355piinsag9dgr...@4ax.com>, David
>> C.
>> Ullrich <dull...@sprynet.com> wrote:
>>
>> > I'm really not sure what the requirement is, but
>> have you condsidered
>> > the standard construction of the completion of a
>> metric space?
>>
>> With his requirements it looks like what he has to do
>> is this ( done
>> with the example of the rationals ) ... Take the
>> usual completion of
>> the rationals, then remove the points that correspond
>> to rationals and
>> replace them with the original rationals.
>
>So that, consider for a metric space M the standard construction of its completion, say M',
>together with the canonical embedding j : M --> M'.
>Next, define F(M) := M \/ (M' \ j(M)) (disjoint union), endowed with a natural metric d.
>[Namely, this metric preserves the original metric of M, the metric on M' \ j(M) inherited from
>the constructed one on M', and d(m,m') = d_{M'}(j(m),m') if m in M and m' in M' \ j(M).]
>It should works. Thank you.
First, why in the world do you think this is a more "natural"
completion than just using M'.
Second: Note that you used the "disjoint union". In fact it
_could_ happen that M and M' \ j(M) are not disjoint,
so you're going to have to "decorate" the sets somehow
to get the disjoint union - hence your "natural" completion
doesn't _really_ contain M as a subset _anyway_!
Technicality: If A and B are sets a standard version of
the "disjoint union" is
{(x, 0) : x in A} union {(x,1) : x in B}.
If that's what you mean by "disjoint union" then what
I said: you're not getting a completion that contains
M anyway.
If you had some other construction of a "disjoint
union" in mind, exactly what is it? Of course there
are ways to do it, but we want to see something
"natural". Given two sets A and B, their disjoint
union is
A union _what_?
>
>> Seems "less natural" to me.
>
>Why should be the above construction "less natural" than identifying an element q in Q with
>the equivalence class of rational sequences converging to q (i.e., the equivalence class containing
>the constant sequence (q, q, q, ...) ) ?
Suppose for example you want to _prove_ something about the
completion, for example prove that it's a completion of M.
With your version anything you prove about x, y in M' is
going to split into four cases (two of which will be equivalent
by symmetry).
>P.S. A weak point of your suggestion : one cannot have "multiplicativity", i.e.
>F(M_1 x M_2) = F(M_1) x F(M_2), that is desirable.
David C. Ullrich
Marc Olschok
Perhaps the "=" above actually means "isomorphic".
Then the conditions certainly make sense.
--
Marc
omega
Let's say it's a question of taste.
>
> Second: Note that you used the "disjoint union". In
> fact it
> _could_ happen that M and M' \ j(M) are not disjoint,
Example needed. In my view, M and M' \ j(M) are like Q and R \ Q, respectively.
[I.e., the "original elements" and the "extra elements"]
> so you're going to have to "decorate" the sets
> somehow
> to get the disjoint union - hence your "natural"
> completion
> doesn't _really_ contain M as a subset _anyway_!
So that you are claiming the answer to the OP is negative ?
BTW, why not "decorate" M' \ j(M) only ?
>
> Technicality: If A and B are sets a standard version
> of
> the "disjoint union" is
>
> {(x, 0) : x in A} union {(x,1) : x in B}.
Then who is the disjoint union of A := Q, and B := R \ Q ?
Not related question : is R a part of C, or not ? Your
definition is the standard one, but is not unique.
>
> If that's what you mean by "disjoint union" then what
> I said: you're not getting a completion that contains
> M anyway.
It follows that, in your opinion, any completion of a metric space I have,
it cannot contain the metric space itself.
And the completion of a Banach space is *not* that Banach space.
>
> If you had some other construction of a "disjoint
> union" in mind, exactly what is it? Of course there
> are ways to do it, but we want to see something
> "natural". Given two sets A and B, their disjoint
> union is
>
> A union _what_?
>
> >
> >> Seems "less natural" to me.
> >
> >Why should be the above construction "less natural"
> than identifying an element q in Q with
> >the equivalence class of rational sequences
> converging to q (i.e., the equivalence class
> containing
> >the constant sequence (q, q, q, ...) ) ?
>
> Suppose for example you want to _prove_ something
> about the
> completion, for example prove that it's a completion
> of M.
> With your version anything you prove about x, y in M'
> is
> going to split into four cases (two of which will be
> equivalent
> by symmetry).
This is not so embarassing as saying a rational number is in fact not a rational number,
but merely an equivalence class of Cauchy sequences.
And the three cases naturally arise when dealing with both rationals and irrationals.
>
> >P.S. A weak point of your suggestion : one cannot
> have "multiplicativity", i.e.
> >F(M_1 x M_2) = F(M_1) x F(M_2), that is desirable.
Nevertheless, despite the above construction (which is not my idea, and I'm not very happy with it), could you provide me with a yes/no answer to my OP ? [My feeling is the answer is negative : there is no such completion functor F : Met --> CMet, preserving the complete metric spaces. But this is to be proved.]
omega
In fact, your comments make very little sense here.
But I think this is your standard answer when facing a correctly stated mathematical yes/no problem that you are not able to solve
>
> -- m
Martin Michael Musatov
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
All of the below is harvested http://www.meami.org/?cx=000961116824240632825%3A5n3yth9xwbo&cof=FORID%3A9%3B+NB%3A1&ie=UTF-8&q=yes+or+no+problem+p%3Dnp+musatov#1102
I don't have to get too fancy here for you to see a pattern and potential. An above average intelligent person with a big imagination should be able to see as much in it.
Musatov Prime Generalization Conjecture/Melded to Prove P==NP ...Jul 1, 2009 ... string, and outputs "yes" or "no". If there is an algorithm (say a ... every problem in NP is reducible to them (under some notion of ...
Math Forum DiscussionsConceptually, a decision problem is a problem that takes as input some string, and outputs "yes" or "no". If there is an algorithm (say a .... Read, Re: Musatov Prime Generalization Conjecture/Melded to Prove P==NP ...
Musatov Prime Generalization Conjecture - sci.math.num-analysis ...more than a polynomial number of strings of length n), then P = NP. ... Conceptually, a decision problem is a problem that takes as input some string, and outputs "yes" or "no". If there is an algorithm (say a ...
Math Forum DiscussionsRe: Musatov Prime Generalization Conjecture/Melded to Prove P==NP Posted: Jul 1, 2009 9:30 PM .... string, and outputs "yes" or "no". If there is an algorithm (say a ... every problem in NP is reducible to them (under some notion of ...
Re: Musatov Prime Generalization ConjectureJul 2, 2009 ... Yes. Proof: Suppose that P = 2n. If n > 1, then P is not a prime, since P is not equal to 2 and is divisible by 2. ... The 3n + 1 problem ... If n is even, divide by 2. If n is odd, ... And if we could resolve P versus NP it would mean we'd be off to the ... whose makespan is no greater than . ...
P=NP Proof Published at CERN...you have no means to refute my proof P==NP based on the knowledge ... I don't see how that follows, but yes, he thinks he's solved ... [ABOVE IS MARTIN MICHAEL MUSATOV'S P==NP PROOF TEXT POSED AS A ... deny it by the evidence shown it solves several NP-complete problems referenced here:STP (STEINER TREE PROBLEM) ...
www.groupsrv.com/science/post-3048142.html
Re: "Musatov"...Yes, of course. You are above all and you are. with. no fault and no ..... science called. P=NP. It is not by _any means_ a tiny little problem. ...
The Awakening of The American Mind: P=NP Incomplete___The Maximal ...... and in fact I did, thereby answering question, yes, I did. ... 7) Therefore, a consistent and complete statement of the problem of P=NP is possible. ... Given this case, P==NP is no longer an eventuality, it is therefore a certainty. ... binary solution to the P=NP problem was formulated by Martin M. Musatov. ...
Re: "Musatov"...Yes, of course. You are above all and you are with no fault and no .... over a tiny little problem in theoretical computer science called. P=NP. ...
Math Forum DiscussionsYes. > > > Proof: Suppose that P = 2n. If n > 1, then P is not a prime, since P is not equal to 2 and is divisible by 2. ... The 3n + 1 problem ... If n is even, divide by 2. If n is odd, ... whose makespan is no greater than . .... Read, Re: Musatov Prime Generalization Conjecture/Melded to Prove P==NP ...
--
MMM (Musatov)
Mariano Suárez-Alvarez
No, the problem makes little sense because in the study
of metric spaces the "nature" of the points is irrelevant.
It is trivial to construct a functor F from metric spaces
to complete metric spaces such that F(X) = X, an *equality*,
not an isomorphism, when X is complete.
Let G(X) be the usual completion, as constructed in every
textbook from Cauchy sequences, and let i : X --> G(X)
the the canonical injection from X to that completion.
Now consider the cartesian product K(X) = G(X) x {X},
endowed with the obvious metric, so that the first
projection K(X) --> G(X) is an isometry; this step is just
to make sure that K(X) and X are disjoint. Now define F(X)
to be, as a set, the union of X and the complement
of i(X) x {X} in K(X), and let j : X --> F(X) be the
inclusion. There is a unique way of defining a metric on X
in such a way that the function j is an isometry into its image,
and that the obvious map F(X) --> G(X) is also an isometry.
This defines F : Met --> CMet on objects. Anyone with the
sufficient amount of motivation can see how to define it
on morphisms of Met.
Now it is obvious that if X is a metric space which
happens to be complete, then F(X) coincides with X
and not only it is isometric to it. You can trivially check
that your condition 1 is satisfied.
Any question which has this stupid construction as a
solution is pretty uninteresting...
-- m
omega
What about 2) ? It seems that your stupid construction is not a solution...
David C. Ullrich
One example where they're disjoint proves that they're always
disjoint? Wow.
>
>> so you're going to have to "decorate" the sets
>> somehow
>> to get the disjoint union - hence your "natural"
>> completion
>> doesn't _really_ contain M as a subset _anyway_!
>
>So that you are claiming the answer to the OP is negative ?
>
>BTW, why not "decorate" M' \ j(M) only ?
Exactly _how_ do you do that in general, in a way
that guarantees it's disjoint from M?
You ignored my question about this. So I'll ask
again: Exactly _how_ can you define "disjoint
union" in such a way that the disjoint union
of A and B is "A union something"?
>>
>> Technicality: If A and B are sets a standard version
>> of
>> the "disjoint union" is
>>
>> {(x, 0) : x in A} union {(x,1) : x in B}.
>
>Then who is the disjoint union of A := Q, and B := R \ Q ?
>Not related question : is R a part of C, or not ? Your
>definition is the standard one, but is not unique.
>
>>
>> If that's what you mean by "disjoint union" then what
>> I said: you're not getting a completion that contains
>> M anyway.
>
>It follows that, in your opinion, any completion of a metric space I have,
>it cannot contain the metric space itself.
What? That's ridiculous. It's ridiculous, and no it does not follow
from what I said. Of course any metric space has a completion
that contains the original set.
omega
And where the heck I claimed this ? Please don't put your words into my mouth.
I just asked for an example of "non-disjointness", and you ignored this question.
>
> >
> >> so you're going to have to "decorate" the sets
> >> somehow
> >> to get the disjoint union - hence your "natural"
> >> completion
> >> doesn't _really_ contain M as a subset _anyway_!
> >
> >So that you are claiming the answer to the OP is
> negative ?
> >
> >BTW, why not "decorate" M' \ j(M) only ?
>
> Exactly _how_ do you do that in general, in a way
> that guarantees it's disjoint from M?
Well, Mariano did. And he did it well. If you don't read all postings in this thread, let me quote from Mariano :
"Let G(X) be the usual completion, as constructed in every
textbook from Cauchy sequences, and let i : X --> G(X)
the the canonical injection from X to that completion.
Now consider the cartesian product K(X) = G(X) x {X},
endowed with the obvious metric, so that the first
projection K(X) --> G(X) is an isometry; this step is just
to make sure that K(X) and X are disjoint. Now define F(X)
to be, as a set, the union of X and the complement
of i(X) x {X} in K(X), and let j : X --> F(X) be the
inclusion. There is a unique way of defining a metric on X
in such a way that the function j is an isometry into its image,
and that the obvious map F(X) --> G(X) is also an isometry."
Satisfied ?
Tim Little
> If you had some other construction of a "disjoint
> union" in mind, exactly what is it? Of course there
> are ways to do it, but we want to see something
> "natural". Given two sets A and B, their disjoint
> union is
>
> A union _what_?
A u B u { x in A /\ B : (x, A u B) } seems to be at least somewhat
natural, certainly more so than the disjoint union constructed with
(a,0) & (b,1) pairs.
Neither A nor B can contain any element of the form (x, A u B) as that
would violate the axiom of regularity, so the elements constructed
with the ordered pairs are all distinct from the elements of A u B.
It's "natural" in the sense that it doesn't involve arbitrary sets
such as 0 or 1, it contains all the elements of the usual union, and
it doesn't maintain a distinction between the elements from A and
those from B. Only the elements in the intersection are distinguished,
by having an extra "copy" in the ordered pairs that augment the usual
union.
- Tim