Local compactness and compactification

Definition: Local compactness

A topological space $X$ is locally compact at a point $x\in X$ if there exists a compact subspace $C\subseteq X$ and a neighborhood $U$ of $x$ such that $U\subseteq C$. We say $X$ is locally compact if this holds for all $x\in X$.

Compactification is the process of adding “points at infinity” to turn an arbitrary space compact so that nice theorems, such as

Definition: Compactification

A compactification of a topological space $x$ is a pair $(X’,i)$ of a compact topological space $X’$ and an embedding $i:X\to X’$ such that $j(X)$ is a dense subset of $X’$, meaning every point of $X’$ is either in $X$ or a limit point of $X$.

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One-point compactification §

Theorem: One-point compactification

A topological space $X$ is locally compact and Hausdorff if and only if there exists a space $Y$ and an embedding $i:X\to Y$ such that:

• (i) $Y$ is compact and Hausdorff;
• (ii) $Y\backslash i(X)=\{p\}$, a single point;
• (iii) $(Y,i)$ is unique, in the sense that if any other $(Y’,i’)$ satisfies these properties, there exists a unique homeomorphism $f:Y\to Y’$ such that $f\circ i=i’$ (someday…commutative algebra diagram).

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Examples §

• Every compact set is locally compact.
• $\mathbb{R}$ is locally compact, since any $x\in \mathbb{R}$ is contained in a compact subset $C=[x-1,x+1]$, for example, and the neighborhood $(x-1,x+1)\subseteq C$. Similarly, every ${\mathbb{R}}^n$ is compact.
• The countably infinite product of the real line with itself with the p

$\mathbb{Q}$ is not locally compact §

$\mathbb{Q}$ is not locally compact as a subset of $\mathbb{R}$ with the usual topology. First, for any irrational $x\notin \mathbb{Q}$ with a neighborhood $(-ϵ,ϵ)\subset \mathbb{R}$, we may find a sequence of rationals in $[-ϵ,ϵ]$ that converges that converges to $x$, which implies that $[-ϵ,ϵ]\cap \mathbb{Q}$.

Now suppose, towards a contradiction, that there exists an open neighborhood $U\subseteq \mathbb{Q}$ of $0$ such that $U\subseteq K\subseteq \mathbb{Q}$ for some compact set $K$. Then there exists a basic open neighborhood $(-ϵ,ϵ)\cap \mathbb{Q}\subseteq U$, which implies that $[-ϵ,ϵ]\cap \mathbb{Q}$ is a closed subset of the compact set $K$, and is therefore compact. But this contradicts the previous paragraph, so we conclude that $0$ has no open neighborhood contained in a compact subset $K\subseteq \mathbb{Q}$.