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

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William Elliot

Posts: 2,432
Registered: 1/8/12
Re: Onto [0,1]
Posted: Apr 27, 2013 4:04 AM
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> 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?

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