Re: automorphisms of subspaces of the reals

On Thu, 21 Jul 2005 08:21:19 -0500, David C. Ullrich
<ullrich@xxxxxxxxxxxxxxxx> wrote:

>On 20 Jul 2005 21:44:37 -0700, "Butch Malahide"
><fred.galvin@xxxxxxxxx> wrote:
>
>>
>>
>>matthias@xxxxxxxxxxx wrote:
>>> Q. Does there exist an infinite subspace X of the reals (with the
>>> relative topology on X), such that the only automorphism of X is the
>>> identity.
>>>
>>> A. I don't believe so. Here is a proof using the Cantor--Bendixson
>>> analysis of subsets of the reals.
>>>
>>> Let U be an infinite subset of the reals. By Cantor--Bendixson, we can
>>> write U as the union of a perfect closed set and a countable set.
>>
>>The Cantor-Bendixson theorem applies to *closed* sets. How do you write
>>the set of all irrational numbers as the union of a perfect closed set
>>and a countable set?
>>
>>> [. . .]
>>>
>>> The proof actually shows that every infinite set of reals has
>>> infinitely many autohomeomorphisms.
>>
>>Actually, assuming the axiom of choice, there is a dense subset U of
>>the real line such that the only continuous mapping f:U-->U is the
>>identity.

Hmm - as you point out this is nonsense. For a second I assumed
that the problem was with the detail I said I couldn't do below,
but no, that detail is clear, there's an error elsewhere, that
gets fixed if we assume that f is injective:

>>Hint: Use transfinite induction. To start with, you can assume that U
>>contains the set Q of all rational numbers. Then, any continuous
>>mapping f:U-->R will be determined by its restriction to Q. There are
>>just continuum many continuous (or otherwise) mappings from Q into R.
>>In constructing the set U, make sure that none of those functions
>>(except the identity) extends to a continuous mapping of U into U.
>
>Ah. I thought there should be a counterexample, but I couldn't see
>how to construct one explicitly. What you say here seems probably
>right - having thought about it for about a minute I'm stuck on
>one detail (maybe you can fill in the detail or say how the thing
>should be approached differently so as to avoid the problem):
>
>If f : Q -> R is continuous say B(f) is the set of all real x
>such that either f has no limit at x or f has a limit at x
>but this limit does not equal x. In general say _f(x) is the
>limit of f at x if there is such a limit, or 0 if the limit
>does not exist.
>
>It seems clear that if f : Q -> R is continuous but not the
>identity then B(f) has cardinality c; this is the detail I
>don't quite see how to prove, not that I've tried very hard.

As you point out, this is very easy to see - there's an interval
I such that f(Q intersect I) is disjoint from I, and hence
I is contained in B(f).

>Assuming this:
>
>Say the continuous functions f : Q -> R other than the
>identity are "enumerated" as f_a for ordinals a < c.
>We "construct" two increasing families of sets U_a and
>T_a for ordinals a < c, such that U_a intersect T_a is
>empty for all a, as follows:
>
>Let U_0 = Q and T_0 = {}.
>
>For each a, choose x in B(f) such that x is not in
>U_b for any b < a, and such that _if_ f has a limit at
>x then _f(x) is not in T_b for any b < a.

The error is here - if, say, f is constant then there's
only one possible _f(x), so we can't choose x so that
_f(x) is not in T_b for b < a.

But if f is continuous and injective and I is as
above then f(Q intersect I) has no isolated points,
so its closure has cardinality c, and we're set.

>Let U_a be
>the union of the previous U_b's with x added, and
>let T_a be the union of the previous T_b's, with
>_f(x) added if f has a limit at x.
>
>The fact that B(f) has cardinality c says that it's
>always possible to find such an x. Now let U be the
>union of the U_a for a < c and let T be the union of
>the T_a.
>
>Now if f : U -> U is continuous and not the identity
>then there exists a such that the restriction of f to
>Q is f_a. Say x = x_a is the real that was added to U_a
>above. Now the fact that f is continuous and x is in
>U shows that f_a has a limit y at x, and that f(x) = y.
>But this is a contradiction, since the construction
>shows that y is in T, hence not in U.
>
>
>************************
>
>David C. Ullrich


************************

David C. Ullrich
.

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