(Theorem) 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).

Proof from Topology.

The reverse implication is easier to show:

1. Show $X$ is Hausdorff. Restricting the embedding $i:X\to Y$ to the domain $i(X)\subseteq Y$ gives a homeomorphism. Since $Y$ is Hausdorff, the subspace $i(X)$ is also Hausdorff, and $i(X)\cong X$ implies $X$ is Hausdorff.
2. Show $X$ is locally compact. Let $x\in X$. Since $p$ is the complement of $X$ in $Y$, there exist disjoint neighborhoods $U$ of $i(x)$ and $V$ of $p\in Y$. Then the set $C=Y\backslash V$ is a closed subset of $i(X)$, and hence compact. Now note that $i^{-1}(U)$ is a neighborhood of $x$ and $i^{-1}(C)$ is compact since the embedding $i$ is a homeomorphism. Since $U\subseteq C$ implies $i^{-1}(U)\subset i^{-1}(C)$, we know $X$ is locally compact at $x$. This holds for any $x\in X$, so $X$ is locally compact overall.

For the forward implication:

1. Define a topology on $Y$. We declare open sets to be of two types: (1) any open subset $U\subset X$, and (2) $Y\backslash C$ where $C\subseteq X$ is compact. This is a topology because:
   • Finite intersections. We need to consider three cases:
     • (1) intersecting (1) produces another type (1) open set;
     • (1) intersecting (2) produces a type (1) open set. Explicitly, $U\cap (Y\backslash C)=U\cap (X\backslash C)$, and $X$ Hausdorff $⟹C$ closed $⟹$ $X\backslash C$ open;
     • (2) intersecting (2) produces another type (2), since $(Y\backslash C_1)\cap (Y\backslash C_2)=Y\backslash (C_1\cup C_2)$.
   • Arbitrary unions.
     • (2) union with (2) produces another (2). Explicitly, ${\bigcup }_{\alpha }(Y\backslash C_{\alpha })=Y\backslash ({\bigcap }_{\alpha }{C}_{\alpha })$. We know each $C_{\alpha }$ closed, since compact subspaces of Hausdorff spaces are closed $⟹$ ${\bigcap }_{\alpha }{C}_{\alpha }$ closed $⟹{\bigcap }_{\alpha }{C}_{\alpha }$ compact since closed subspaces of compact spaces are compact.
     • (1) union with (2) produces (2). Note that arbitrary unions of (1)s and (2)s have already been shown to be (1), (2) respectively, so suffices to consider union of a single type (1) with a single type (2). We have $U\cup (Y\backslash C)=Y\backslash (C\backslash U)$, and $C\backslash U$ is compact because $X$ is Hausdorff.
2. Check the subspace topology on $X\subseteq Y$ is the same as the topology on $X$.
   • Type (1) open sets: $U\cap X=U$ is clearly open in $X$.
   • Type (2) open sets: $(Y\backslash C)\cap X=X\backslash C$ is open in $X$.
3. Show $Y$ compact using compactness of $X$. Let $\mathcal{C}$ be open cover of $Y$. Then there exists a type (2) set $Y\backslash C$ where $C$ is compact. Then $C$ is covered by all type (1) sets, and we can find a finite subcover $\{U_1,\dots ,U_n\}$. Then $Y$ has the finite cover $\{U_1,\dots ,U_n,Y\backslash C\}$.
4. Show $Y$ Hausdorff using local compactness of $X$. Let $x\in X$. Choose a compact set $C\subseteq X$ and an open neighborhood $U$ of $x$ such that $U\subset C$. For any other point $p\in Y$, suppose $p\notin X$ (we know $X$ is Hausdorff, so this is uninteresting). Then $Y\backslash C$ is an open neighborhood of $p$, and we have $U\cap (Y\backslash C)$ for open neighborhoods of $x\ne p$, respectively.