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From: Tim Little on 26 Jul 2010 22:29
On 2010-07-26, William Elliot <marsh(a)rdrop.remove.com> wrote:
> It's not rhetorical! There's an unique example.
> Let S = {p,q} with the topology { nulset, {p}, S }.

In that example, why is the function {p->q, q->q} not continuous?


- Tim
From: William Elliot on 26 Jul 2010 23:20
On Mon, 27 Jul 2010, Tim Little wrote:

> On 2010-07-26, William Elliot <marsh(a)rdrop.remove.com> wrote:
>> It's not rhetorical! There's an unique example.
>> Let S = {p,q} with the topology { nulset, {p}, S }.
>
> In that example, why is the function {p->q, q->q} not continuous?
>
It is. This thread has self destructed.
Expect a reincardination soon.
From: William Elliot on 26 Jul 2010 23:35
On Mon, 26 Jul 2010, [ISO-8859-1] Jos� Carlos Santos wrote:
> 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_.

Ohhh, as the late night fog dissipates,
this thread is now taken off life support.

I have a vision of a revision
that requires a little incision,
to revive for a final decision.
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