```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 verticallines {x}xC, x in C. (I'm identifying the cardinal number c with thecorresponding initial ordinal.)Let (H_n: n < c) enumerate the homeomorphs of C in CxC (or theuncountable closed subsets of CxC, or the uncountable Borel sets, orthe uncountable analytic sets; the point is that there are only c ofthem, 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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