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Re: Onto [0,1]
Posted:
Apr 26, 2013 12:07 AM


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.



