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From: William Elliot on 26 Jul 2010 06:45
On Mon, 26 Jul 2010, Bill wrote:
> Butch Malahide wrote:
>> On Jul 26, 4:25 am, William Elliot

>>> Let S be a space and p a point in S.
>>>
>>> What topology can S have if a function f:S -> S
>>> is continuous iff f(p) = p?
>>> Is the topology unique?
>>
>> Inasmuch as constant functions are continuous, p must be the only point
>> in the space. Yes, the topology on a one-point space is unique.
>
> I think the OP means that f is the identity.

I do not. I didn't say for all p. I picked a specific point p.

> The case |S|=1 appears to be settled.

Trivial.

> What happens if |S|=2 (rhetorical)?

It's not rhetorical! There's an unique example.
Let S = {p,q} with the topology { nulset, {p}, S }.

When S has more than two points, I find the topology isn't unique.

Thus the refined puzzle.
Does S have a unique T0 topology with the given property?

From: David Hartley on 26 Jul 2010 06:54
In message <i2jo2m0fo1(a)news6.newsguy.com>, Bill <Bill(a)NOSPAM.net> writes
>Butch Malahide wrote:
>> On Jul 26, 4:25 am, William Elliot<ma...(a)rdrop.remove.com> wrote:
>>> Let S be a space and p a point in S.
>>>
>>> What topology can S have if a function f:S -> S
>>> is continuous iff f(p) = p?
>>>
>>> Is the topology unique?
>>
>> Inasmuch as constant functions are continuous, p must be the only
>> point in the space. Yes, the topology on a one-point space is unique.
>
>I think the OP means that f is the identity. The case |S|=1 appears to
>be settled. What happens if |S|=2 (rhetorical)?

Perhaps the OP meant f to be onto - i.e. the identity is the only
continuous mapping of S onto itself.

For any well-ordering on S, the set of initial segments provides such a
topology. Any infinite S has many non-isomorphic well-orderings, so the
topology is not unique.
--
David Hartley
From: Bill on 26 Jul 2010 07:36
William Elliot wrote:

>> What happens if |S|=2 (rhetorical)?
>
> It's not rhetorical! There's an unique example.

Sorry, I only meant that I did not pose the question
as one for myself, but instead in an effort to (try to) be helpful.
You are evidently on the right track.

Bill
From: Bill on 26 Jul 2010 07:41
William Elliot wrote:
> On Mon, 26 Jul 2010, Bill wrote:
>> Butch Malahide wrote:
>>> On Jul 26, 4:25 am, William Elliot
>
>>>> Let S be a space and p a point in S.
>>>>
>>>> What topology can S have if a function f:S -> S
>>>> is continuous iff f(p) = p?
>>>> Is the topology unique?
>>>
>>> Inasmuch as constant functions are continuous, p must be the only
>>> point in the space. Yes, the topology on a one-point space is unique.
>>
>> I think the OP means that f is the identity.
>
> I do not. I didn't say for all p. I picked a specific point p.
>
>> The case |S|=1 appears to be settled.
>
> Trivial.
>
>> What happens if |S|=2 (rhetorical)?
>
> It's not rhetorical! There's an unique example.
> Let S = {p,q} with the topology { nulset, {p}, S }.
>
> When S has more than two points, I find the topology isn't unique.
>
> Thus the refined puzzle.
> Does S have a unique T0 topology with the given property?
>

What happens if |S|=3? : )

From: José Carlos Santos on 26 Jul 2010 08:36
On 26-07-2010 11:31, William Elliot wrote:

>>> Let S be a space and p a point in S.
>>>
>>> What topology can S have if a function f:S -> S
>>> is continuous iff f(p) = p?
>>>
>>> Is the topology unique?
>>
>> Every constant function is continuous. So your condition means that S
>> has no point other than p.
>>
> False. See my reply to Bill.

No, it is not false. Suppose that your space has some other point _q_
besides _p_. The constant function _f_ which always takes the value _q_
is continuous, for whatever topology you take on S. But f(p) is
different from _p_. Therefore, your condition cannot hold if there is
some point in S different from _p_.

Best regards,

Jose Carlos Santos
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