Date: Apr 26, 2013 12:07 AM
Author: Butch Malahide
Subject: Re: Onto [0,1]
On Apr 25, 9:50 pm, quasi <qu...@null.set> wrote:

> Butch Malahide wrote:

> >>quasi wrote:

> >> > Prove or disprove:

>

> >> > If X is a topological space and f: X -> [0,1] is a

> >> > continuous surjection, then X has a subspace homeomorphic

> >> > to the Cantor set.

>

> >> [. . .]

> >Here's a less trivial (because it's nonconstructive)

> >counterexample. Let C be the Cantor set.

> >Let c = |C| = 2^{aleph_0}. The product space CxC contains just

> >c subsets homeomorphic to C; of course, each of those subsets

> >has cardinality c. By transfinite induction, we can construct

> >a subset X of CxC which meets each vertical line {x}xC (x in

> >C) while containing no homeomorph of C. Of course there is a

> >continuous surjection from X to C.

>

> How do you ensure that X contains no homeomorph of C?

Same way you construct a Bernstein set.

Let (L_n: n < c) be a transfinite sequence enumerating the vertical

lines {x}xC, x in C. (I'm identifying the cardinal number c with the

corresponding initial ordinal.)

Let (H_n: n < c) enumerate the homeomorphs of C in CxC (or the

uncountable closed subsets of CxC, or the uncountable Borel sets, or

the uncountable analytic sets; the point is that there are only c of

them, and each of them has cardinality c).

At step n, choose x_n in L_n\{y_m: m < n}, and then choose y_n in H_n\

{x_m: m <= n}.

Finally, let X = {x_n: n < c}. Note that y_n is in H_n\X.