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quasi
Posts:
12,042
Registered:
7/15/05


Re: Onto [0,1]
Posted:
Apr 26, 2013 1:04 AM


Butch Malahide wrote: >quasi 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.
Nice.
Thank you.
quasi



