Re: automorphisms of subspaces of the reals

Butch Malahide wrote:
> Butch Malahide wrote:
> >
> > 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.
>
> Nonsense! For one thing, there are the *constant* functions f:U-->U.
> Maybe I meant to say that the only continuous *injection* from U into U
> is the identity. Let me think this over.

Let's see if I can get it right this time. Let R be the real line, let
c = |R| be the cardinal of the continuum, and let Q be the set of all
rational numbers.

PROPOSITION. Assuming the axiom of choice, there is a dense subset U of
R such that the only continuous injection from U to U is the identity.

We construct U by transfinite induction. At a typical stage in the
construction, we have defined two disjoint subsets X,Y of R, each of
cardinality less than c; X is the set of points we have decided to put
in U, Y the set of points we have decided not to put in U; and X
contains Q.

Now consider a continuous injection f:Q-->R, not the identity. We want
to "spoil" f by adding a few points to X and Y so as to guarantee that
f can not be the restriction of a continuous injection from U to U.

Define g:R-->R so that g(x) is equal to the limit of f at x whenever
the limit exists, otherwise we don't care.

Since f is continuous but not the identity, there is an open interval I
of R such that, for all rational q in I, f(q) is not in I.

Let A = {x in I\Y: g(x) is in X}, and B = {x in I\Y: g(x) is not in X}.

If B is nonempty, choose x in B; add x to X, and add g(x) to Y.

If B is empty, then |A| = c > |X|; choose distinct points x_1 and x_2
in A with g(x_1) = g(x_2); add x_1 and x_2 to X.

.

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