continuity of extended real valued function
sto
extended real numbers. Assume that f(x)=oo for all x. Then establishing
the continuity of f as a function on a metric space has the difficulty
that, for any x and y in X, the quantity
|f(x) - f(y)|
is undefined because the subtraction of two infinities is not defined. Can
I correctly argue that f is continuous because the preimage of any open set
is either {} or X, both of which are open sets?
Thanks
-sto
William Elliot
Yes. Exercise: show every constant valued function
between any two topological spaces, is continuous.
sto
But the problem is that to consider f as a map between two metric
spaces, rather than two topological spaces, it is necessary to put a
metric on the extended reals. I don't see how a metric can be put on
the extended reals in the presence of +oo and -oo; and without a metric,
how do you put a topology on the extended reals?
thanks,
-sto
David C. Ullrich
Well, if you don't know what the topology on the extended reals
is then you shouldn't be asking how to show a given function is
continuous, you need to first ask what it _means_ to say
that function is continuous! With no topology in sight
"continuous" is meaningless - you need to know what
a statement _means_ before thinking about how to prove it.
(Your post sounds like a homework problem. There _should_
be a definition of the topology somewhere in the book;
maybe you missed it.)
Anyway. There's a standard topology on the extended reals.
A basis at +oo is given by the sets (a, +oo]; similarly for
-oo.
This topology _is_ metrizable, so you can just talk about
convergent sequences. And it turns out that x_j -> x in
this topology if and only if x_j -> x in the sense you
learned in calculus.
Not that it matters much, but you can define a metric
on the extended reals like so:
If x and y are both finite,
d(x,y) = |x-y| / (1 + |x-y|).
If one or both of x and y is infinite, then
d(x,y) = 0 if x = y and d(x,y) = 1 otherwise.
DU.
>thanks,
>-sto
Ilmari Karonen
> On Sat, 19 Mar 2011 11:04:54 -0400, sto <s...@address.invalid> wrote:
>>
>> I don't see how a metric can be put on the extended reals in the
>> presence of +oo and -oo; and without a metric, how do you put a
>> topology on the extended reals?
>
> on the extended reals like so:
>
> If x and y are both finite,
>
> d(x,y) = |x-y| / (1 + |x-y|).
>
> If one or both of x and y is infinite, then
> d(x,y) = 0 if x = y and d(x,y) = 1 otherwise.
Or, my personal favorite,
d(x,y) = |arctan(x) - arctan(y)|,
where arctan(oo) = pi/2 and arctan(-oo) = -pi/2.
Or course, the arctangent is only one possible choice: the inverse of
any bounded monotone continuous function R -> R, suitably extended at
+/- oo, will do just as well (for both mine and David's constructions,
in fact).
--
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.
sto
not true? If it is, then any constant function should be continuous.
> (Your post sounds like a homework problem. There _should_
> be a definition of the topology somewhere in the book;
> maybe you missed it.)
>
I don't have a topology book to look in.
My post is not a homework problem. I am working through the proof that
if (X,d) is a metric space and (X,S,m) is a measure space in which every
measurable set can be approximated from above by an open set and below
by a closed set, then the continuous functions are dense in L1(X,m).
Briefly, since the simple functions are dense in L1, it is sufficient to
show that the continuous functions are dense in the simple functions.
But any simple function is a finite linear combination of characteristic
functions of measurable sets, so it is sufficient to show that the
continuous functions are dense in the characteristic functions of
measurable sets. Given any measurable set A and any e>0, pick an open
set G and closed set F having the property that F subset A subset G and
m(G\F) < e. Then define the function
h(x) = D( X\G, x) / ( D( X\G, x) + D(F,x) )
where D is defined, for any set S, by
D( S, x) = inf {d(x,y) : y in S}
For a given nonempty set S, I can show that D is a continuous function
of x, in the sense of a map between the metric space X and the reals (as
a metric space). It follows that so is the function h. Furthermore the
L1 norm || h - I_A || < e, where I_A is the characteristic function of
A. This completes the proof.
But look at the function D. It is continuous in x for any *nonempty*
set S, which is sufficient for the proof. But the book (translated from
Russian) seems to say that D is continuous for any (arbitrary) set S.
Is that statement true? If S is empty, then D( S,x) = oo for all x. Is
D(S,x)=oo a continuous function and if so in what sense of continuity?
David C. Ullrich
>On 3/19/11 11:19 AM, David C. Ullrich wrote:
>> On Sat, 19 Mar 2011 11:04:54 -0400, sto<s...@address.invalid> wrote:
>>
>>> On 3/18/11 11:40 PM, William Elliot wrote:
>>>> On Fri, 18 Mar 2011, sto wrote:
>>>>
>>>>> Let X be a metric space and f be a function on X taking values in the
>>>>> extended real numbers. Assume that f(x)=oo for all x. Then establishing
>>>>> the continuity of f as a function on a metric space has the difficulty
>>>>> that, for any x and y in X, the quantity
>>>>>
>>>>> |f(x) - f(y)|
>>>>>
>>>>> is undefined because the subtraction of two infinities is not defined.
>>>>
>>>>> Can I correctly argue that f is continuous because the preimage of any
>>>>> open set is either {} or X, both of which are open sets?
>>>>>
>>>> Yes. Exercise: show every constant valued function
>>>> between any two topological spaces, is continuous.
>>> I did that above: the preimage of any open set is either empty or X.
>>> But the problem is that to consider f as a map between two metric
>>> spaces, rather than two topological spaces, it is necessary to put a
>>> metric on the extended reals. I don't see how a metric can be put on
>>> the extended reals in the presence of +oo and -oo; and without a metric,
>>> how do you put a topology on the extended reals?
Did you bother to read my post?
>> Well, if you don't know what the topology on the extended reals
>> is then you shouldn't be asking how to show a given function is
>> continuous, you need to first ask what it _means_ to say
>> that function is continuous! With no topology in sight
>> "continuous" is meaningless - you need to know what
>> a statement _means_ before thinking about how to prove it.
>>
>I was speculating that {} and X were open sets in any topology. Is that
>not true? If it is, then any constant function should be continuous.
>
>> (Your post sounds like a homework problem. There _should_
>> be a definition of the topology somewhere in the book;
>> maybe you missed it.)
>>
>I don't have a topology book to look in.
>My post is not a homework problem. I am working through the proof that
>if (X,d) is a metric space and (X,S,m) is a measure space in which every
>measurable set can be approximated from above by an open set and below
>by a closed set, then the continuous functions are dense in L1(X,m).
This should perhaps be phrased a little differently, since for example
the continuous functions need not even be _elements_ of L^1.
Usually "A is dense in B" entails that A is a subset of B.
>Briefly, since the simple functions are dense in L1, it is sufficient to
>show that the continuous functions are dense in the simple functions.
>But any simple function is a finite linear combination of characteristic
>functions of measurable sets, so it is sufficient to show that the
>continuous functions are dense in the characteristic functions of
>measurable sets. Given any measurable set A and any e>0, pick an open
>set G and closed set F having the property that F subset A subset G and
>m(G\F) < e. Then define the function
>
> h(x) = D( X\G, x) / ( D( X\G, x) + D(F,x) )
>
>where D is defined, for any set S, by
>
> D( S, x) = inf {d(x,y) : y in S}
>
>For a given nonempty set S, I can show that D is a continuous function
>of x, in the sense of a map between the metric space X and the reals (as
>a metric space). It follows that so is the function h. Furthermore the
>L1 norm || h - I_A || < e, where I_A is the characteristic function of
>A. This completes the proof.
Oh? Evidently there's something in the proof that requires that S be
nonempty - you should find that.
>But look at the function D. It is continuous in x for any *nonempty*
>set S, which is sufficient for the proof. But the book (translated from
>Russian) seems to say that D is continuous for any (arbitrary) set S.
>Is that statement true? If S is empty, then D( S,x) = oo for all x. Is
>D(S,x)=oo a continuous function and if so in what sense of continuity?
Reagarding whether that function is continuous: When you post
a question and someone answers it, you might consider reading
the reply before repeating the question.
Regarding that theorem - the author was talking about
real-valued (or complex-valued) continuous functions.
If S is empty then his construction deoesn't work - the
function which is identially zero is continuous, but
it's not in L^1, and is certainly not an approximation
to the characteristic function of the empty set.
The author was silently ignoring trivialities - if
S is empty then the characteristic function of S
_is_ continuous, so there's nothing to prove.
So he was assuming that S is nonempty.
sto
> On Tue, 22 Mar 2011 00:47:56 -0400, sto<s...@address.invalid> wrote:
>
>> On 3/19/11 11:19 AM, David C. Ullrich wrote:
>>> On Sat, 19 Mar 2011 11:04:54 -0400, sto<s...@address.invalid> wrote:
>>>
>>>> On 3/18/11 11:40 PM, William Elliot wrote:
>>>>> On Fri, 18 Mar 2011, sto wrote:
>>>>>
>>>>>> Let X be a metric space and f be a function on X taking values in the
>>>>>> extended real numbers. Assume that f(x)=oo for all x. Then establishing
>>>>>> the continuity of f as a function on a metric space has the difficulty
>>>>>> that, for any x and y in X, the quantity
>>>>>>
>>>>>> |f(x) - f(y)|
>>>>>>
>>>>>> is undefined because the subtraction of two infinities is not defined.
>>>>>
>>>>>> Can I correctly argue that f is continuous because the preimage of any
>>>>>> open set is either {} or X, both of which are open sets?
>>>>>>
>>>>> Yes. Exercise: show every constant valued function
>>>>> between any two topological spaces, is continuous.
>>>> I did that above: the preimage of any open set is either empty or X.
>>>> But the problem is that to consider f as a map between two metric
>>>> spaces, rather than two topological spaces, it is necessary to put a
>>>> metric on the extended reals. I don't see how a metric can be put on
>>>> the extended reals in the presence of +oo and -oo; and without a metric,
>>>> how do you put a topology on the extended reals?
>
> Did you bother to read my post?
>
>>> Well, if you don't know what the topology on the extended reals
>>> is then you shouldn't be asking how to show a given function is
>>> continuous, you need to first ask what it _means_ to say
>>> that function is continuous! With no topology in sight
>>> "continuous" is meaningless - you need to know what
>>> a statement _means_ before thinking about how to prove it.
>>>
>> I was speculating that {} and X were open sets in any topology. Is that
>> not true? If it is, then any constant function should be continuous.
>>
>>> (Your post sounds like a homework problem. There _should_
>>> be a definition of the topology somewhere in the book;
>>> maybe you missed it.)
>>>
>> I don't have a topology book to look in.
>> My post is not a homework problem. I am working through the proof that
>> if (X,d) is a metric space and (X,S,m) is a measure space in which every
>> measurable set can be approximated from above by an open set and below
>> by a closed set, then the continuous functions are dense in L1(X,m).
>
> This should perhaps be phrased a little differently, since for example
> the continuous functions need not even be _elements_ of L^1.
> Usually "A is dense in B" entails that A is a subset of B.
>
This answers another question I had. I found the phrasing "continuous
functions are dense in L1(X,m)" confusing, since f(x) = |x| on the real
line with Lebesque measure is continuous and not in L1, but that's how
the theorem was phrased in the book. I would conjecture that "any
element of L1(X,m) can be approximated (in L1) by a continuous function"
is more accurate.
>> Briefly, since the simple functions are dense in L1, it is sufficient to
>> show that the continuous functions are dense in the simple functions.
>> But any simple function is a finite linear combination of characteristic
>> functions of measurable sets, so it is sufficient to show that the
>> continuous functions are dense in the characteristic functions of
>> measurable sets. Given any measurable set A and any e>0, pick an open
>> set G and closed set F having the property that F subset A subset G and
>> m(G\F)< e. Then define the function
>>
>> h(x) = D( X\G, x) / ( D( X\G, x) + D(F,x) )
>>
>> where D is defined, for any set S, by
>>
>> D( S, x) = inf {d(x,y) : y in S}
>>
>> For a given nonempty set S, I can show that D is a continuous function
>> of x, in the sense of a map between the metric space X and the reals (as
>> a metric space). It follows that so is the function h. Furthermore the
>> L1 norm || h - I_A ||< e, where I_A is the characteristic function of
>> A. This completes the proof.
>
> Oh? Evidently there's something in the proof that requires that S be
> nonempty - you should find that.
>
I was assuming that a simple function had to take at least one non-zero
value on a set of non-zero measure, in which case S would have to be
non-empty. But of course, that leaves out the function f(x) = 0.
It does not say in the book that S has to be non-empty. It is unclear
whether S is assumed to be nonempty or whether the proof is supposed to
hold for any S, whether empty or not. But if S is empty, then D takes
on infinite values, which is unusual. Of course, it also not stated
what the range of D is.
>> But look at the function D. It is continuous in x for any *nonempty*
>> set S, which is sufficient for the proof. But the book (translated from
>> Russian) seems to say that D is continuous for any (arbitrary) set S.
>> Is that statement true? If S is empty, then D( S,x) = oo for all x. Is
>> D(S,x)=oo a continuous function and if so in what sense of continuity?
>
> Reagarding whether that function is continuous: When you post
> a question and someone answers it, you might consider reading
> the reply before repeating the question.
>
It's not as simple as that. I read your post (twice), but I am
simultaneously trying to guess the definition of the function D (its
domain and range are not precisely stated anywhere) and trying to prove
whether my guess can be called "continuous" in some sense of that word.
So I have to guess what I need to prove and prove it at the same time.
> Regarding that theorem - the author was talking about
> real-valued (or complex-valued) continuous functions.
> If S is empty then his construction deoesn't work - the
> function which is identially zero is continuous, but
> it's not in L^1, and is certainly not an approximation
> to the characteristic function of the empty set.
>
> The author was silently ignoring trivialities - if
> S is empty then the characteristic function of S
> _is_ continuous, so there's nothing to prove.
> So he was assuming that S is nonempty.
>
It had not occurred to me to consider the empty set as a separate case,
and I think that is the critical issue, not the continuity of D.
I think that explains everything. Without actually saying so, the
author proves only the case for non-empty measurable sets. Although it
does not say so, the proof does not actually apply to the empty set;
that has to be considered as a separate case. That means the function D
is real valued, its argument is always a non-empty set, and its
continuity is straightforward.