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By spaces I mean locally compact, $\sigma$-compact, connected, locally connected, Hausdorff topological spaces.

I need a reference (not a proof, I already have one -- or at least I think so; counterexamples are welcome) for the following fact. I believe it should be standard.

If $X$ is a non-compact space, then the Alexandroff one-point compactification of $X$ is homeomorphic to the quotient space $\mathfrak{F}X\big/\mathcal{E}(X)$, where $\mathfrak{F}X$ is the Freudenthal compactification of $X$ and $\mathcal{E}(X)$ is the set of ends of $X$.

Definitions of both $\mathfrak{F}X$ and $\mathcal{E}(X)$ are recalled below.

Recall that an end of a topological space $X$ is a function $$\varepsilon: \{K\subseteq X: K \text{ is compact }\} \to 2^X\smallsetminus\{\emptyset\}$$ such that $\varepsilon(K)$ is an unbounded (= of non-compact closure) connected component of $X\smallsetminus K$ and $\varepsilon(L)\subseteq \varepsilon(K)$ for all compact sets $K\subseteq L \subseteq X$. The set of ends of $X$ is denoted by $\mathcal{E}(X)$.

For $\varepsilon \in \mathcal{E}(X)$ and a compact set $K\subseteq X$ define $$N(\varepsilon,K)=(X\smallsetminus K) \cup \bigl\{\delta\in\mathcal{E}(X):\delta(K)=\varepsilon(K)\bigr\}.$$

Let $\mathfrak{F}X=X\cup \mathcal{E}(X)$ with topology given by the following bases of open neighbourhoods: for a basis of neighbourhoods of $x\in X$ we take a basis of neighbourhoods of $x$ in $X$; for $\varepsilon\in\mathcal{E}(X)$ the basis of neighbourhoods is $$\{N(\varepsilon,K):K\subseteq X \text{ is compact }\}.$$

Michał Kukieła
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1 Answers1

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There is nothing special about the Freudenthal compactification in this respect. This holds for any (Hausdorff) compactification of any non-compact locally compact Hausdorff space. As a reference you might use Engelking's General Topology, Theorem 3.5.11.

THE ALEXANDROFF COMPACTIFICATION THEOREM. Every non-compact locally compact space X has a compactification $\omega X$ with one-point remainder. This compactification is the smallest element in $C(X)$ with respect to the order $\le$, its weight is equal to the weight of the space X.

$C(X)$ refers to the collection of equivalence classes of compactifications of $X$ and $\le$ is defined by

Let $c_2X \le c_1X$ if there exists a continuous mapping $f: c_1X \to c_2X$ such that $fc_1 = c_2$;

Note that any such $f$ is a quotient map.

Niels J. Diepeveen
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