Thread: one point compactification of the positive integers

1. 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.

2. Originally Posted by math8
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 $\displaystyle \mathbb{N}$ consists of $\displaystyle \mathbb{N}$ together with a single point which we can call $\displaystyle \infty$. The topological structure is that of the discrete topology on $\displaystyle \mathbb{N}$; and the open neighbourhoods of $\displaystyle \infty$ are by definition the complements of the compact subsets of $\displaystyle \mathbb{N}$. The compact subsets of $\displaystyle \mathbb{N}$ are the finite subsets, so the neighbourhoods of $\displaystyle \infty$ are the sets with finite complement.

The map $\displaystyle n\mapsto 1/n$ (where $\displaystyle 1/\infty$ is interpreted as 0) takes this space to the set $\displaystyle K=\{0\} \cup \{1/n : n \in\mathbb{N}\}$, 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).

3. Originally Posted by math8
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 $\displaystyle f:X \rightarrow Y$ 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 $\displaystyle \mathbb{Z_+}$ and p be a single point of $\displaystyle Y \setminus \mathbb{Z_+}$.

We define a bijective function $\displaystyle f:Y \rightarrow Z$ such that f(n) = 1/n where $\displaystyle n \in \mathbb{Z_+}$ 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 $\displaystyle K=\{1/n : n \in \mathbb{Z_+}\}$ is open in Z.
2. All sets of the form $\displaystyle Z \setminus C$, where C is a compact subspace of $\displaystyle K$, 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 $\displaystyle x \in K$ and $\displaystyle y = 0$, then we can choose a compact set C in $\displaystyle K$ containing a neighborhood U of x. Then U and $\displaystyle Z \setminus C$ 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 $\displaystyle \mathbb{Z_+}$, the inverse image of an open set in K under f is open in $\displaystyle Y \setminus \{p\} \subset Y$.
Let U be a neighborhood of Z containing 0. U contains 0 and all but finite points of K. Let $\displaystyle P = \{p_1,p_2,...,p_N\}$ be the finite points in K not contained in U. Now, the inverse image of an open set U under f is

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

We see that $\displaystyle f^{-1}(U)$ 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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