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From: William Elliot on 12 Jul 2010 04:03
If f is a continuous function on a compact space X,
(sj)_j a sequence within X and (f(sj))_j -> y, is
there some x with f(x) = y?

Would addition of hypothesis such as Hausdorff, normal, 1st
countable (or even metric if need be) ease or settle the issue?


From: David C. Ullrich on 12 Jul 2010 05:29
On Mon, 12 Jul 2010 01:03:01 -0700, William Elliot
<marsh(a)rdrop.remove.com> wrote:

>If f is a continuous function on a compact space X,
>(sj)_j a sequence within X and (f(sj))_j -> y, is
>there some x with f(x) = y?

Any net in X has a convergent subnet.

Or: Let E_n be the closure of the set of s_j with j > n.
Consider the intersection of the E_n.

>Would addition of hypothesis such as Hausdorff, normal, 1st
>countable (or even metric if need be) ease or settle the issue?
>

From: William Elliot on 12 Jul 2010 06:32
On Mon, 12 Jul 2010, David C. Ullrich wrote:
> <marsh(a)rdrop.remove.com> wrote:
>
>> If f is a continuous function on a compact space X,
>> (sj)_j a sequence within X and (f(sj))_j -> y, is
>> there some x with f(x) = y?
>
> Any net in X has a convergent subnet.
>
Isn't that over kill? Anyway, that's the key.

Some a in X and subsequence t of s = (sj)_j with t -> a
f.t = (f(tj))_j -> f(a). Since ft is a subsequence of fs
ft -> y and by assuming codomain is Hausdorff, f(a) = y.

Do inverse limits have anything to do with this?

> Or: Let E_n be the closure of the set of s_j with j > n.
> Consider the intersection of the E_n.
>
It's not empty, nor need it be a singleton.
Does f(/\_n E_n) = {y} ? No.

Let X = [0,1], codomain Y = [0,2] with cofinite topology,
sequence s be a listing of rationals in [0,1], f the identity.
Of course fs -> any a in [0,2], however f(/\_n E_n) = [0,1].
So if fs -> 2, there's no x with f(x) = y.

Assuming Y is Hausdorff, does f(/\_n E_n) = {y}?

>> Would addition of hypothesis such as Hausdorff, normal, 1st
>> countable (or even metric if need be) ease or settle the issue?
From: William Elliot on 13 Jul 2010 06:34
On Mon, 12 Jul 2010, David C. Ullrich wrote:
> <marsh(a)rdrop.remove.com> wrote:
>
>> If f is a continuous function on a compact space X,
>> (sj)_j a sequence within X and (f(sj))_j -> y, is
>> there some x with f(x) = y?
>
> Any net in X has a convergent subnet.
>
The complete and exact theorem is as below.

If f is a continuous function from a sequentially compact space X into
a Hausdorff space Y, (sj)_j a sequence within X and (f(sj))_j -> y,
then there's some x with f(x) = y.

The proof is nearly immediate by picking x to
be the limit of some subsequence of (sj)_j.

> Or: Let E_n be the closure of the set of s_j with j > n.
> Consider the intersection of the E_n.
>
From: David C. Ullrich on 13 Jul 2010 08:30
On Mon, 12 Jul 2010 03:32:57 -0700, William Elliot
<marsh(a)rdrop.remove.com> wrote:

>On Mon, 12 Jul 2010, David C. Ullrich wrote:
>> <marsh(a)rdrop.remove.com> wrote:
>>
>>> If f is a continuous function on a compact space X,
>>> (sj)_j a sequence within X and (f(sj))_j -> y, is
>>> there some x with f(x) = y?
>>
>> Any net in X has a convergent subnet.
>>
>Isn't that over kill?

I guess, if posting a hint that gives a "one-line"
solution to a simple problem is "overkill".
(In any case, the fact that you might be
unfamiliar with nets is why I gave the second
hint...)

>Anyway, that's the key.
>
>Some a in X and subsequence t of s = (sj)_j with t -> a

No such subsequence. Subnet.

>f.t = (f(tj))_j -> f(a). Since ft is a subsequence of fs
>ft -> y and by assuming codomain is Hausdorff, f(a) = y.
>
>Do inverse limits have anything to do with this?
>
>> Or: Let E_n be the closure of the set of s_j with j > n.
>> Consider the intersection of the E_n.
>>
>It's not empty, nor need it be a singleton.
>Does f(/\_n E_n) = {y} ? No.
>
>Let X = [0,1], codomain Y = [0,2] with cofinite topology,
>sequence s be a listing of rationals in [0,1], f the identity.
>Of course fs -> any a in [0,2], however f(/\_n E_n) = [0,1].
>So if fs -> 2, there's no x with f(x) = y.

Yes, it's clear that the answer to your question must be no
if we don't assume anything about Y.

>Assuming Y is Hausdorff, does f(/\_n E_n) = {y}?

Yes. This is very easy.

Suppose x is in the intersection of the E_n, and let f(x) = y'.

Suppose that y' <> y. Choose disjoint open sets O and O'
with y in O and y' in O'.

Since f(x_j) -> y, there exists N such that f(x_j) is in O
for all j > N. Since f is continuous it follows that
f(E_N) is contained in the closure of O, and hence
that y' is not in f(E_N), which contradicts the fact
that f(x) = y'.



>>> Would addition of hypothesis such as Hausdorff, normal, 1st
>>> countable (or even metric if need be) ease or settle the issue?

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