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

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quasi

Posts: 10,208
Registered: 7/15/05
Re: Onto [0,1]
Posted: Apr 26, 2013 1:04 AM
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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



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