Theorem 4 (Alexandroff)

If $X$ is any locally compact space, there exists a compact space $X'$ and a homeomorphism $f$ of $X$ onto the complement of a point in $X'$. Furthermore, if $X'_1$ is another compact space such that there is a homeomorphism $f_1$ of $X$ onto the complement of a point in $X'_1$, then there is a unique homeomorphism $g$ of $X'$ onto $X'_1$ such that $f_1 = g \circ f$.

Let us begin by proving the second assertion of the theorem. Let

If the homeomorphism $g$ exists it must be unique, for by definition we have $g(x') = f_1(\overline{f}(x'))$ if $x' \neq \omega$ and therefore $g(\omega) = \omega_1$. It remains to show that the bijection $g : X' \to X'_1$ thus defined is bicontinuous; since $X'$ and $X'_1$ are interchangeable we need only show that the image under $g$ of a neighbourhood of a point $x' \in X'$ is a neighbourhood of $g(x')$ in $X'_1$. This is obvious from the definition of $g$ if $x' \neq \omega$. If $x' = \omega$, let $V'$ be an open neighbourhood of $\omega$ in $X'$; then $X' - V' = K$ is closed in $X'$ and therefore compact (no. 3, Proposition 3) and is contained in $f(X)$; hence $g(K) = f_1(\overline{f}(K))$ is compact (no. 4, Theorem 2, Corollary 1). It follows that $g(V') = X'_1 - g(K)$ is an open neighbourhood of $\omega_1$ (no. 3, Proposition 4). Hence $g$ is a homeomorphism.

To prove the first part of the theorem, let $X'$ be a set which is the sum of $X$ and a set consisting of a single point $\omega$. We define a topology on $X'$ by taking the set $\mathscr{O}$ of open subsets of $X'$ to consist of all open subsets of $X$ and all subsets of the form $(X - K) \cup \{ \omega \}$, where $K$ is a compact subset of $X$. Since any intersection of compact subsets of $X$ is compact (no. 3, Propositions 3 and 4) and since any closed subset of a compact set is compact (no. 3, Proposition 3), it follows that $\mathfrak{D}$ satisfies axiom $(O_I)$; and since any finite union of compact subsets of $X$ is compact (no. 3, Proposition 5), $\mathfrak{D}$ also satisfies $(O_{II})$. Every compact subset of $X$ is closed in $X$ (no. 3, Proposition 4) and therefore the topology induced on $X$ by that of $X'$ is the original topology on $X$. Thus it remains to show that $X'$ is compact. In the first place, $X'$ is Hausdorff. For if $x, y$ are any two distinct points of $X$, they have disjoint open neighbourhoods $V, W$ respectively in $X$, and $V, W$ are open in $X'$; on the other hand each $x \in X$ has a compact neighbourhood $K$ in $X$, which is also a neighbourhood of $x$ in $X'$, while

is a neighbourhood of $\omega$ in $X'$ and manifestly does not meet $K$. Finally, $X'$ is quasi-compact. Let $(U_\lambda)_{\lambda \in L}$ be an open covering of $X'$; then for at least one index $\mu \in L$ we have $U_\mu = (X - K_\mu) \cup \{ \omega \}$ where $K_\mu$ is a compact subset of $X$. Hence there is a finite subset $H$ of $L$ such that the sets $U_\lambda$ (for $\lambda \in H$) cover $K_\mu$; if $J = H \cup \{ \mu \}$, then $(U_\lambda)_{\lambda \in J}$ is a finite open covering of $X'$, and the proof is complete.

Notice that if $X$ is already compact, then $\omega$ is an isolated point of the compact space $X'$; hence $X'$ is the sum ($\S$ 2, no. 4, Example 3) of the space $X$ and the space $\{ \omega \}$.

When a compact space $X'$ has been constructed as above from a locally compact space $X$ by adjoining an element $\omega$, it is often said that $\omega$ is the "point at infinity" of $X'$, and that $X'$ is obtained from $X$ by adjoining a point at infinity. $X'$ is also called the Alexandroff compactification or the one-point compactification of the locally compact space $X$.

\* Example. If we apply Alexandroff's theorem to the real plane $\mathbf{R}^2$, we get a compact space homeomorphic to the sphere $S_2$ whose equation is $x_1^2 + x_2^2 + x_3^2 = 1$ in $\mathbf{R}^3$. A homeomorphism of these two spaces may be described as follows: the point $\omega$ (the point at infinity) adjoined to $\mathbf{R}^2$ is mapped to $(0, 0, 1) \in S_2$, and every point $(x_1, x_2)$ of $\mathbf{R}^2$ is mapped to the point where the line joining the points $(0, 1, 1)$ and $(x_1, x_2, 0)$ in $\mathbf{R}^3$ meets $S_2$ again. This homeomorphism is known as stereographic projection.