Re: automorphisms of subspaces of the reals



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.

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.

.

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