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Math Help - one point compactification of the positive integers

  1. #1
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    one point compactification of the positive integers

    How do we show the one point compactification of the positive integers is homeomorphic to the set K={0} U {1/n : n is a positive integer}?

    Say Y is the one point compactification of the positive integers. I know Y must contain Z+ and Y\Z+ is a single point. Also Y is a compact Hausdorff space.

    But I am not sure how to show this.
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  2. #2
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    Quote Originally Posted by math8 View Post
    How do we show the one point compactification of the positive integers is homeomorphic to the set K={0} U {1/n : n is a positive integer}?

    Say Y is the one point compactification of the positive integers. I know Y must contain Z+ and Y\Z+ is a single point. Also Y is a compact Hausdorff space.

    But I am not sure how to show this.
    The one-point compactification of \mathbb{N} consists of \mathbb{N} together with a single point which we can call \infty. The topological structure is that of the discrete topology on \mathbb{N}; and the open neighbourhoods of \infty are by definition the complements of the compact subsets of \mathbb{N}. The compact subsets of \mathbb{N} are the finite subsets, so the neighbourhoods of \infty are the sets with finite complement.

    The map n\mapsto 1/n (where 1/\infty is interpreted as 0) takes this space to the set K=\{0\} \cup \{1/n : n \in\mathbb{N}\}, and you can check that it preserves the topology (because the points 1/n are all isolated, and the neighbourhoods of 0 are exactly the sets with finite complement).
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  3. #3
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    Quote Originally Posted by math8 View Post
    How do we show the one point compactification of the positive integers is homeomorphic to the set K={0} U {1/n : n is a positive integer}?

    Say Y is the one point compactification of the positive integers. I know Y must contain Z+ and Y\Z+ is a single point. Also Y is a compact Hausdorff space.

    But I am not sure how to show this.
    What Opalg showed is perfectly fine. This is just a detailed approach.

    First of all, I rename K={0} U {1/n : n is a positive integer} as Z and denote K as {1/n : n is a positive integer} for the sake of notational convention.

    Lemma 1. Let f:X \rightarrow Y be a bijective continuous function. If X is compact and Y is Hausdorff, then f is a homeomorphism.

    Let Y be the one-point compactification of \mathbb{Z_+} and p be a single point of Y \setminus \mathbb{Z_+}.

    We define a bijective function f:Y \rightarrow Z such that f(n) = 1/n where n \in \mathbb{Z_+} and f(p) = 0. f is a clearly bijective function.

    By the definition of a one-point compactification, Y is compact. We shall show that Z is Hausdorff in order to apply lemma 1. We topologize Z as follows:
    1. Any subset of (isolated) points in K=\{1/n : n \in \mathbb{Z_+}\} is open in Z.
    2. All sets of the form Z \setminus C, where C is a compact subspace of K, are open in Z.

    Since K is locally compact Hausdorff space, if two points lie in K, it satisfies the Hausdorff condition automatically. Meanwhile, if x \in K and y = 0, then we can choose a compact set C in K containing a neighborhood U of x. Then U and Z \setminus C are disjoint neighborhoods of x and y, respectively, in Z. Thus, Z is Hausdorff.

    We remain to show that f is continuous.
    Since the inverse of an isolated point in K under f is an isolated point in \mathbb{Z_+}, the inverse image of an open set in K under f is open in Y \setminus \{p\} \subset Y.
    Let U be a neighborhood of Z containing 0. U contains 0 and all but finite points of K. Let P = \{p_1,p_2,...,p_N\} be the finite points in K not contained in U. Now, the inverse image of an open set U under f is

    f^{-1}(U) = \{p\} \text{ }\cup \text{ } \mathbb{Z_+}\setminus\{c_1, c_2, ... , c_N \}, where  f(c_k) = p_k, k=1,2,...,N.

    We see that f^{-1}(U) is open in Y because its complement is compact in Y.Thus, f is continuous.

    It follows that f is a required homeomorphism by lemma 1.
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