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




Re: Onto [0,1]
Posted:
Apr 27, 2013 4:04 AM


> Correctly: every *nonempty* compact metric space is a continuous image > of the Cantor set. (Likewise, every nonempty separable complete metric > space is a continuous image of the space of irrational numbers.) > > Let Y be a nonempty compact metric space. Then, for some natural n_1, > Y is the union of n_1 nonempty closed sets of diameter < 1. Next, for > some natural n_2, each of those n_1 sets is the union of n_2 (not > necessarily distinct) nonempty closed sets of diameter < 1/2. Next, > for some natural n_3, each of the previously chosen n_1*n_2 sets is > the union of n_3 nonempty closed sets of diameter < 1/n. And so on.
Cover Y with { B(y,1)  y in Y }. Thus Y is the union of n_1 sets of the form cl B(y,1) with not empty interior.
For n_2, since 1/2 is making this proof too long, cover each of the n_1 compact sets K, with cl(K /\ B(y,1/2^2)), etc.
One notes the K's can be constructed to be not empty regular closed sets
> Use these coverings in the obvious way to define a continuous > surjection from the infinite product space X = D(n_1)xD(n_2)x. . . to > Y, where D(n) is a discrete space of cardinality n.
Nothing obvious at all about it. f:X > Y, x > to the unique element of a nest of compact sets as constructed above. It's a notational nightmare.
Since such a nest can be constructed for each y in Y, f is surjective.
Pointwise continuity of f, though seemingly possible, isn't apparent for the mess of details. Is there another way to show f is continuous?
> Finally, observe that X is a continuous image (in fact a homeomorph but > we don't need that) of the Cantor set.
Every not empty perfect, zero dimensional, 2nd countable, compact Hausdorff space is homeomorphic to {0,1}^2.
So {0,1}^N is homeomorphic to {0,1,.. n}^N, n in N?
Is your proof an example of an inverse limit topology?



