One-Point Compactification of a Topological Space

One-Point Compactification of a Topological Space

Consider a topological space $(X, \tau)$. Then $X$ may or may not be compact depending on the topology $\tau$. If $X$ is not compact, then we might want to construct a compact topological space from $X$ while preserving as much of the topology $\tau$ on $X$ as possible. We formally define this process below.

Definition: Let $(X, \tau)$ be a topological space that is not compact. A Compactification $(Y, \tau')$ of $X$ is a topological space such that $X$ is a subspace of $Y$ and such that $X$ is dense in $Y$, i.e., $\bar{X} = Y$.

We begin by looking at perhaps the simplest type of compactification called the one-point compactification of a topological space.

Definition: Let $(X, \tau)$ be a topological space that is not compact. A One-Point Compactification is the topological space $(X_{\infty}, \tau')$ is the set $X_{\infty} = X \cup \{ \infty \}$ (where $\infty$ denotes any point not in $X$) with the topology $\tau'$ given as $\tau' = \tau \cup \{ (X \setminus C) \cup \{ \infty \} : C \: \mathrm{is \: compact \: and \: closed \: in \:} X \}$.

This definition might seem a bit strange, but let's look at an easier-to-visualize example. Consider the real line $\mathbb{R}$ and the unit circle with the point $(0, 1)$ removed:

\begin{align} \quad S^1 \setminus \{ (0, 1) \} = \{ (x, y) : x^2 + y^2 = 1 \} \setminus \{ (0, 1) \} \end{align}

Then $S^1 \setminus \{ (0, 1) \}$ is homeomorphic to $\mathbb{R}$ through what are known as the stereographic projection map $f : S^1 \setminus \{ (0, 1) \} \to \mathbb{R}$ defined for all $(x, y) \in S^1 \setminus \{ (0, 1) \}$ by:

\begin{align} f(x, y) = \frac{x}{1 - y} \end{align}

For each point $(x, y) \in S^1 \setminus \{ (0, 1) \}$ the value of $f(x, y)$ will be the intersection of the line that passes through $(0, 1)$ and $(x, y)$ with the $x$-axis as illustrated below:


If we think of the point $\infty$ as being the point at the ends of the number line (equivalent to the missing pole $(0, 1)$) then we can construct a one-point compactification of $\mathbb{R}$ as $(\mathbb{R}_{\infty} \tau')$. In other words, $\mathbb{R}_{\infty} = \mathbb{R} \cup \{ \infty \}$, and the open sets in $\mathbb{R}_{\infty}$ are of two types.

The first type of open sets in $\tau'$ are sets that are normally open in $\mathbb{R}$:


The second type of open sets in $\tau'$ are the complements of closed and compact sets in $\mathbb{R}$:

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