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From: David C. Ullrich on 13 Jul 2010 08:32
On Tue, 13 Jul 2010 03:34:55 -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.
>>
>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 result is true for compact spaces as well as for sequentially
compact spaces. This sort of thing is the whole point to nets.

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

Or "subnet" for compact X.

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

From: William Elliot on 14 Jul 2010 04:43
On Tue, 13 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 result is true for compact spaces as well as for sequentially
> compact spaces. This sort of thing is the whole point to nets.
>
>> The proof is nearly immediate by picking x to
>> be the limit of some subsequence of (sj)_j.
>
> Or "subnet" for compact X.
>
Ok, since sequences are nets, the stronger theorem is

compact X, Hausdorff Y, net s into X, fs -> y ==> some x with f(x) = y.

fs is of course, the composition of two functions f and s:N -> X,
if s is a sequence or s:L -> X, if s is a net where L is the
directed preorder domain of the net s.

Subnets are such a nuisance I was hoping to avoid them.
Oh well, may as well learn up on them. Somehow nets
seem easier at times, than filters.

Besides swallowing the cofinal definition of
subnet, these are the theorems I need to prove.

If s is a net into a compact space then s has a cluster point.
Each cluster point of a net has a subnet converging to it.
If s is a subnet of a net that converges to x, then s -> x.

That's quite an homework assignment.
How long do I have to complete the lesson?

----
From: William Elliot on 14 Jul 2010 06:20
On Tue, 13 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.

>> 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'.
>
Nice, avoids filters and nets. However the technique you used
looks like it'd work to show every net into a compact set has
a cluster point.

If s:L -> S is a net into S and S is compact,
for all j in L, let Kj = cl{ sk | j <= k }.
{ Kj | j in L } has finite intersection property as L is directed.
Thus by compactness, some x in /\_j Kj. If open U nhood x,
then for all j, x in Kj, U /\ { sk | j <= k } nonnul, showing
x is a cluster pt of s.

From: William Elliot on 14 Jul 2010 23:49
On Tue, 13 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.
>> Anyway, that's the key.
>> Some a in X and subsequence t of s = (sj)_j with t -> a
>
> No such subsequence. Subnet.
>
Here's simpler without subnets.

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

Since X is compact, (sj)_j has a cluster point x.
As f is continuous, f(x) is a cluster point of (f(sj))_j.
Since (f(sj))_j -> y and Y is Hausdorff, f(x) = y, QED.

--
For sequences, there's a different theorem with the same proof.
If f is a continuous function on a countably 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.

Thus two different theorems, one for nets, the other for
sequences, and one not a generalization of the other.

----
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