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From: Butch Malahide on 27 Jul 2010 05:13
On Jul 26, 11:13 pm, William Elliot <ma...(a)rdrop.remove.com> wrote:
> Let S be a set and p an element of S.
>
> What topology can S have if for all functions f:S -> S,
> f is continuous iff f(p) = p or f is constant function?
>
> Is the topology unique?  If it's required that the space
> be T0, will the topology be unique?

A topology with the stated property is *necessarily* T0. Assume for a
contradiction that there are two distinct points a,b in S such that
every neighborhood of a is a neighborhood of b and vice versa. Then
every function
f:S -> {a, b} is continuous. Choose c in {a, b} \ {p}, and define
f:S -> {a, b} by setting f(a) = b, f(b) = a, and f(x) = c otherwise.
Then f is a nonconstant continuous selfmap on S, and f(p) is not p.
From: Butch Malahide on 27 Jul 2010 05:21
On Jul 27, 4:13 am, Butch Malahide <fred.gal...(a)gmail.com> wrote:
> On Jul 26, 11:13 pm, William Elliot <ma...(a)rdrop.remove.com> wrote:
>
> > Let S be a set and p an element of S.
>
> > What topology can S have if for all functions f:S -> S,
> > f is continuous iff f(p) = p or f is constant function?
>
> > Is the topology unique?  If it's required that the space
> > be T0, will the topology be unique?
>
> A topology with the stated property is *necessarily* T0. Assume for a
> contradiction that there are two distinct points a,b in S such that
> every neighborhood of a is a neighborhood of b and vice versa. Then
> every function
> f:S -> {a, b} is continuous. Choose c in {a, b} \ {p}, and define
> f:S -> {a, b} by setting f(a) = b, f(b) = a, and f(x) = c otherwise..
> Then f is a nonconstant continuous selfmap on S, and f(p) is not p.

Sorry, I made that too complicated. Just define f(a) = b and f(x) = a
for all x in S \ {a}. Then f is a continuous selfmap with no fixed
points.
From: Butch Malahide on 28 Jul 2010 01:52
On Jul 27, 4:21 am, Butch Malahide <fred.gal...(a)gmail.com> wrote:
> On Jul 27, 4:13 am, Butch Malahide <fred.gal...(a)gmail.com> wrote:
>
>
>
>
>
> > On Jul 26, 11:13 pm, William Elliot <ma...(a)rdrop.remove.com> wrote:
>
> > > Let S be a set and p an element of S.
>
> > > What topology can S have if for all functions f:S -> S,
> > > f is continuous iff f(p) = p or f is constant function?
>
> > > Is the topology unique?  If it's required that the space
> > > be T0, will the topology be unique?
>
> > A topology with the stated property is *necessarily* T0.

The OP's "puzzle" seems obscurely motivated, but this observation
might be an amusing little exercise (hardly a puzzle) for beginning
students of topology.

EXERCISE. If a topological space X has the fixed point property (every
continuous mapping of X into X has a fixed point) then X is a T_0
space.

SOLUTION. Let a and b be two distinct points in X. Define a function
f:X->X by setting f(a) = b, f(x) = a if x is different from a. Since f
has no fixed points, it is discontinuous; i.e., there is an open set U
whose preimage is not open. Thus U is an open set containing one and
only one of the two points a and b.
From: William Elliot on 28 Jul 2010 04:51
Butch Malahide <fred.galvin(a)gmail.com>
>> Let S be a set and p an element of S.
>
>> What topology can S have if for all functions f:S -> S,
>> f is continuous iff f(p) = p or f is constant function?
>
>> Is the topology unique? If it's required that the space
>> be T0, will the topology be unique?

> Given S and p, the only topologies satisfying your condition are:
> (I) the open sets are the empty set and all sets containing p;
> (II) the open sets are S and all sets not containing p.

That's a surprise, dual topologies using the self-dual
definition of continuous.

> These are T_0 topologies. If |S| = 1 they coincide; if |S| = 2 they
> are distinct but homeomorphic; if |S| > 2 they are nonhomeomorphic.

> A topology with the stated property is *necessarily* T0. Assume for a
> contradiction that there are two distinct points a,b in S such that
> every neighborhood of a is a neighborhood of b and vice versa. Then
> every function f:S -> {a, b} is continuous.

> Define f(a) = b and f(x) = a, for all x in S \ {a}.
> Then f is a continuous selfmap with no fixed points.

Excellent. Have you any additional evidence backing your
claim that the include p and exclude p topologies are the
only topologies with the stated continuity property?

Having established that S is T0, there'd be two cases:
some open U with p in U, U proper subset S;
some closed K with p in K, K proper subset S.

For the first case assume p in A. Map A to p
and S\A to a point outside of U. Since
f^-1(U) = A, A is open. Now if B is a not empty,
open set with p not in B, then S is disconnected,
which contradicts the continuity property.

The dual second case is immediate with a dual proof.
Alternatively, for x /= p, map S\x to p and x to
a point outside of K. Since f^-1(K) = S\x is
closed, every point /= p, is open and every
set without p is open, ie every set with p
is closed.

----
From: William Elliot on 28 Jul 2010 05:17
On Tue, 27 Jul 2010, Butch Malahide wrote:
>>> On Jul 26, 11:13�pm, William Elliot <ma...(a)rdrop.remove.com> wrote:
>>
>>>> Let S be a set and p an element of S.
>>
>>>> What topology can S have if for all functions f:S -> S,
>>>> f is continuous iff f(p) = p or f is constant function?
>>
>>>> Is the topology unique? �If it's required that the space
>>>> be T0, will the topology be unique?
>>
>>> A topology with the stated property is *necessarily* T0.
>
> The OP's "puzzle" seems obscurely motivated, but this observation
> might be an amusing little exercise (hardly a puzzle) for beginning
> students of topology.
>
It was motivated by a student's problem, characterize the
continuous maps from Sp to Sq where Sp has the include
p topology and Sq has the include q topology, for S.

> EXERCISE. If a topological space X has the fixed point property (every
> continuous mapping of X into X has a fixed point) then X is a T_0
> space.
>
Exercise. Fixed point spaces are connected.

> SOLUTION. Let a and b be two distinct points in X. Define a function
> f:X->X by setting f(a) = b, f(x) = a if x is different from a. Since f
> has no fixed points, it is discontinuous; i.e., there is an open set U
> whose preimage is not open. Thus U is an open set containing one and
> only one of the two points a and b.
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