Chapter 11: Local Compactness and One-Point Compactification
In the previous chapters, we explored compactness and sequential compactness in topological spaces. In this chapter, we introduce the concept of locally compact spaces and focus on how to turn a non‑compact locally compact space into a compact space by adding the fewest points (a single point) – this is the famous One‑Point Compactification (Alexandroff Compactification). Finally, we will lay some conceptual groundwork for the Urysohn Metrization Theorem.
1. Basic Concepts of Compactification
In topology, we often wish to embed a non‑compact “open” space into a compact space in order to exploit the good properties of compact spaces.
Definition 11.1 (Compactification)
Let \((X, \mathcal{T})\) be a Hausdorff space. A topological space \((Y, \mathcal{T}')\) is called a compactification of \((X, \mathcal{T})\) if it satisfies the following three conditions:
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(1) \(Y\) is a compact Hausdorff space.
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(2) \(X \subseteq Y\), and the subspace topology induced by \(Y\) on \(X\) is exactly \(\mathcal{T}\), i.e., \(\mathcal{T}'|_X = \mathcal{T}\).
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(3) \(X\) is dense in \(Y\), i.e., the closure satisfies:
A simple example: If \((X, \mathcal{T})\) itself is already a compact Hausdorff space, then we can simply take \(Y = X\). In this case, the trivial extension of \(X\) itself is its compactification.
2. One‑Point Compactification (Alexandroff Compactification)
The one‑point compactification is the “smallest” way of compactifying a space, adding only a single point \(\infty\).
Theorem 11.1 (One‑Point Compactification Theorem)
Let \((X, \mathcal{T})\) be a Hausdorff space. The following two statements are equivalent (TFAE):
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(1) \(X\) is locally compact but not compact.
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(2) There exists a one‑point compactification \((Y, \mathcal{T}')\) of \((X, \mathcal{T})\), where
Moreover, in the above situation, if \((Y_1, \mathcal{T}_1)\) and \((Y_2, \mathcal{T}_2)\) are two different one‑point compactifications of \(X\), then there exists a unique homeomorphism between them (and this map is the identity on \(X\)).
Proof of Theorem 11.1 (Click to expand)
Part 1: Uniqueness of the one‑point compactification
Suppose \((Y_1, \mathcal{T}_1)\) and \((Y_2, \mathcal{T}_2)\) are two one‑point compactifications of \(X\), with \(Y_1 = X \cup \{\infty_1\}\), \(Y_2 = X \cup \{\infty_2\}\).
Define a map \(f: Y_1 \to Y_2\) by \(f(x) = x\) for all \(x \in X\), and
Since the map must be the identity on \(X\), such a bijection \(f\) is clearly unique. We only need to show that \(f\) and its inverse are continuous.
Consider an arbitrary open set \(U \subseteq Y_2\) in \(Y_2\):
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Case 1: If \(\infty_2 \notin U\), then \(U \subseteq X\). Since the one‑point compactification preserves the original topology and \(X\) is an open subspace, \(f^{-1}(U) = U\) is also open in \(Y_1\).
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Case 2: If \(\infty_2 \in U\), then its complement \(C = Y_2 \setminus U\) is a closed set in \(Y_2\). Since \(Y_2\) is compact, the closed subset \(C\) is compact, and \(C \subseteq X\). In \(Y_1\), \(f^{-1}(U) = Y_1 \setminus C\). Because \(C\) is compact in \(X\) and \(X\) is Hausdorff, \(C\) is closed in \(Y_1\); hence \(Y_1 \setminus C\) is open in \(Y_1\).
Therefore \(f\) is continuous. By symmetry the same argument shows that \(f^{-1}\) is continuous, so \(f\) is a homeomorphism.
Part 2: Proof of (1) \(\Rightarrow\) (2) (Construction of the one‑point compactification)
Assume that \(X\) is locally compact but not compact. Define \(Y = X \cup \{\infty\}\). We construct a topology \(\mathcal{T}'\) on \(Y\) as follows:
We must verify that \(\mathcal{T}'\) is indeed a topology and satisfies the three conditions of a compactification.
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Verification of the subspace topology: Since \(Y \setminus C\) intersected with \(X\) is \(X \setminus C\), and a compact set \(C\) in a Hausdorff space \(X\) is closed, we have \(\mathcal{T}'|_X = \mathcal{T}\).
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Verification of the Hausdorff property: For two distinct points in \(X\), they can be separated because the original topology is Hausdorff. Now take \(x \in X\) and \(\infty \in Y\): since \(X\) is locally compact, there exists an open set \(U\) containing \(x\) whose closure \(\overline{U}\) in \(X\) is compact.
Let \(V = Y \setminus \overline{U}\); clearly \(V \in \mathcal{T}'\) and \(\infty \in V\). Since \(U \cap V = \emptyset\), we have separated \(x\) and \(\infty\). Thus \(Y\) is Hausdorff.
- Verification of compactness: Let \(\mathcal{A} = \{A_i\}_{i \in I}\) be an arbitrary open cover of \(Y\). There must be some \(A_{i_0}\) that contains \(\infty\).
By the definition of \(\mathcal{T}'\), an open set containing \(\infty\) must be of the form \(A_{i_0} = Y \setminus C\), where \(C\) is a compact subset of \(X\).
For the remaining uncovered set \(C\), since \(C\) is compact, the family of sets \(\{A_i \cap X\}\) from \(\mathcal{A}\) (excluding the one containing \(\infty\)) has a finite subcover for \(C\).
Combining these finitely many open sets with \(A_{i_0}\) gives a finite subcover of \(Y\). Hence \(Y\) is compact.
Part 3: Proof of (2) \(\Rightarrow\) (1)
Assume that a one‑point compactification \(Y = X \cup \{\infty\}\) exists. We show that \(X\) is locally compact and non‑compact.
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Non‑compactness: If \(X\) were compact, then as a compact set in the Hausdorff space \(Y\), it would be closed. Being closed, \(\overline{X} = X \ne Y\), contradicting the density requirement \(\overline{X} = Y\) of a compactification. Therefore \(X\) cannot be compact.
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Local compactness: Take any \(x \in X \subseteq Y\). Since \(Y\) is a compact Hausdorff space (hence normal and can separate points), we can find open sets \(U\) and \(V\) such that
Let \(C := Y \setminus V\). Because \(V\) is open, \(C\) is a closed set in \(Y\).
Since \(Y\) is compact, the closed subset \(C\) is also compact. Clearly, because \(\infty \in V\), we have \(C \subseteq X\).
Now consider the neighbourhood of \(x\): \(x \in U \subseteq \overline{U} \subseteq C\). This shows that \(x\) has an open neighbourhood \(U\) whose closure is compact, which is exactly the definition of local compactness. \(\square\)
3. Local Compactness and Open Subspaces
From the proof of the one‑point compactification theorem, we can extract a very useful equivalent characterisation of locally compact spaces.
Corollary 11.2
A topological space \((X, \mathcal{T})\) is a locally compact Hausdorff space if and only if it is homeomorphic to an open subspace of some compact Hausdorff space.
Proof of Corollary 11.2 (Click to expand)
Necessity (\(\Rightarrow\)):
If \((X, \mathcal{T})\) is a locally compact Hausdorff space, by Theorem 11.1 we can construct its one‑point compactification \(Y = X \cup \{\infty\}\).
In this compact Hausdorff space \(Y\), the singleton \(\{\infty\}\) is compact and therefore closed in a Hausdorff space.
Hence its complement \(Y \setminus \{\infty\} = X\) is an open set in \(Y\). That is, \(X\) embeds as an open subspace of the compact space \(Y\).
Sufficiency (\(\Leftarrow\)):
Let \(Y\) be a compact Hausdorff space and \(X \subseteq Y\) be an open subset of \(Y\). We need to show that the subspace \((X, \mathcal{T}_X)\) is also a locally compact Hausdorff space.
Since the Hausdorff property is inherited by subspaces, \(X\) is obviously Hausdorff.
Now take any \(x \in X\). Since \(Y\) is a compact Hausdorff space (hence also locally compact and regular), there exists an open neighbourhood \(U\) of \(x\) in \(Y\) such that its closure \(cl_Y(U)\) in \(Y\) is compact and
(Because \(X\) is open, we can restrict the neighbourhood of \(x\) to the interior of \(X\)). Since \(cl_Y(U)\) is entirely contained in \(X\), it is also a compact neighbourhood of \(x\) in the topology of \(X\). Therefore \(X\) is locally compact. \(\square\)
4. Conceptual Foundations for the Urysohn Metrization Theorem
Finally, we lay some groundwork for the famous Urysohn Metrization Theorem. This theorem gives a very simple sufficient condition for an abstract topological space to be metrizable (i.e., to admit a metric that induces its topology).
Definition 11.2 (First Countable and Second Countable)
- First Countable: For a topological space \((X, \mathcal{T})\), if for every point \(x \in X\) there exists a countable open neighbourhood base \(\{B_i(x)\}_{i=1}^\infty\). This means that \(x \in B_i(x)\) and, for any open set \(U\) containing \(x\), there exists some \(i\) such that
- Second Countable: If the entire topology \(\mathcal{T}\) possesses a countable topological basis \(\mathcal{B} = \{B_i\}_{i \in \mathbb{N}}\).
Note: Second countability implies first countability, because we can simply take \(\mathcal{B}_x := \{B_i \in \mathcal{B} \mid x \in B_i\}\) as a countable local base at \(x\).
Together with separation axioms, we obtain the core theorem of metrization theory (the complete proof is long; we only state it here):
Theorem 11.3 (Urysohn Metrization Theorem)
If a topological space \((X, \mathcal{T})\) is both second countable and regular, then it is metrizable.