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Re: Onto [0,1]
Posted:
Apr 29, 2013 3:25 PM


On Apr 29, 3:24 am, William Elliot <ma...@panix.com> wrote: > On Sat, 27 Apr 2013, Butch Malahide wrote: > > On Apr 27, 3:04 am, William Elliot <ma...@panix.com> wrote: > > > > 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 < > > > > 1/3. 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. > > > Yes, obviously. [However, I specified sets of *diameter* *less* than > > 1, so change B(y,1) to B(y,1/3).] > > It makes no difference if the diameter of sets is < 1/n or <= 1/n. > What's important is the limit of the diameters is zero. > > > > > > > NOTATION: As usual [n] = {1, 2, 3, ..., n} for natural n. Y is the > > union of nonempty closed sets Y_x (x in [n_1]) of diameter < 1. If x_1 > > in [n_1], x_2 in [n_2], ..., x_{k1} in [n_{k1}], then Y_{x_1, > > x_2, ..., x_{k1}} is the union of nonempty closed sets Y_{x_1, > > x_2, ..., x_{k1}, x_k} (x_k in [n_k]) of diameter < 1/k. > > > > > Use these covengs 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. > > Here is the "mess of details": > > > Suppose x = (x_1, x_2, ...) in X, f(x) = y in Y. Let epsilon > 0 be > > given. Choose n so that 1/n < epsilon. Let U = {u in X: u_i = x_i for > > all i <= n}. Then U is a neighborhood of x in X, and > > f(U) subset Y_{u_1, ..., u_n} = Y_{x_1, ..., x_n} subset B(y, 1/n) > > subset B(y, epsilon), > > Q.E.D. > > > > Finally, observe that X is a continuous image (in fact a homeomorph > > > > but we don't need that) of the Cantor set. > > > Actually, the construction as given allows the possibility that X is > > finite. If we want to ensure that X is homeomorphic to the Cantor set > > C, we can do that by requiring n_i > 1. But there is no need for that. > > Yet it's simpler and more direct than going through C^N.
It's (very slightly) simpler if you don't count the work needed to prove the topological characterization of C, which is harder than what we're proving here. The reader who does not already know that characterization will probably not appreciate your invoking such a difficult result to save an easy twoline argument, and would probably prefer a more selfcontained approach.



