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Example text

So the dimension of X is precisely 2. 7 A nonempty subset S of a 0-dimensional space X is 0dimensional. Proof: The set S clearly has dimension at most 0. It is not empty, so does not have dimension −1. So the dimension must be 0. 38 CHAPTER 1. 8 A nonempty space X has dimension 0 if and only if every point p ∈ X and every closed set C ⊆ X such that p ∈ C can be separated by open sets. Proof: First suppose that X has dimension 0 according to our original definition. Certainly X \ C is a neighborhood of p.

Now define f :S → T (x, y) → t(x, y) x2 + y 2 · (x, y) . It is easy to check that f maps S to T in a one-to-one, onto bicontinuous fashion. 12. The next result is at first a bit surprising. 5 The spaces X = (0, 1) ⊆ R and Y = (0, ∞) are homeomorphic. For let f : X → Y be given by f(x) = x/[1 − x]. Then it is easy to verify that f satisfies all the properties of a homeomorphism of these spaces. This last example illustrates the idea that a homeomorphism does not preserve size or distance. Instead, it preserves the essential topological nature (or the shape) of a space.

If each of the sets Tj , j = 1, 2, . . is nowhere dense in X, then ∪j Tj is nowhere dense in X. In fact there is some classical terminology connected with Baire’s theorem that is worth belaboring (because it is so commonly used, and it aids in one’s understanding).

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Algebraic theories by Wraith, Gavin


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