one point compactification of the positive integers

• Apr 28th 2009, 07:16 AM
math8
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.
• Apr 28th 2009, 12:05 PM
Opalg
Quote:

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).
• Apr 28th 2009, 11:54 PM
aliceinwonderland
Quote:

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.