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Topic: Onto [0,1]
Replies: 40   Last Post: Apr 29, 2013 10:16 PM

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: Onto [0,1]
Posted: Apr 26, 2013 12:07 AM
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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.



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