Chapter 9: Compactness in Euclidean Spaces and Spaces Without Isolated Points
Continuing from the previous chapter's discussion on compactness in general topological spaces, this chapter focuses on the Euclidean space \(\mathbb{R}^n\), which we are most familiar with. We will prove the extremely important Heine–Borel theorem in calculus and derive the extreme value theorem from it. Additionally, we will explore a profound topological property of compact Hausdorff spaces—namely, a compact Hausdorff space without isolated points must be uncountable.
1. Compactness in \(\mathbb{R}^n\) (Heine–Borel Theorem)
In general topological spaces, a compact set is not necessarily closed and bounded, nor is a closed and bounded set necessarily compact. However, in Euclidean space \(\mathbb{R}^n\), the two are equivalent. To prove this core result, we first need to establish that closed intervals on the real line are compact.
Lemma 9.1 (Closed intervals on the real line are compact)
In the real line \(\mathbb{R}\) with the standard topology, any closed interval \([a, b]\) is a compact space.
Proof of Lemma 9.1 (Click to expand)
Let \(\mathcal{A} = \{A_i\}_{i \in I}\) be an open cover of the closed interval \([a, b]\).
Define a set \(S\):
Clearly, since \(a \in [a, b]\) is contained in some open set, we have \(a \in S\). Hence \(S\) is nonempty, and \(S\) is bounded above by \(b\).
By the completeness axiom of the real numbers, \(S\) has a supremum. Let
Claim 1: \(c \in S\).
Since \(c \in [a, b]\) and \(\mathcal{A}\) covers \([a, b]\), there exists an open set \(A_0 \in \mathcal{A}\) with \(c \in A_0\). Because \(A_0\) is open, there exists a point \(d < c\) such that the half‑open interval \((d, c] \subseteq A_0\). Since \(c\) is the supremum of \(S\), there must be a point \(y \in S\) in \((d, c]\) (otherwise \(d\) would be a smaller upper bound). Because \(y \in S\), the subinterval \([a, y]\) can be covered by finitely many open sets from \(\mathcal{A}\). Taking the union of those finitely many open sets with \(A_0\) gives a finite subcover of \([a, c]\). Hence \(c \in S\).
Claim 2: \(c = b\).
Suppose, for contradiction, that \(c < b\). By Claim 1, \(c \in S\) and \(c \in A_0\). Since \(A_0\) is open and \(c < b\), there exists a point \(e\) with \(c < e \le b\) such that the whole interval \([c, e) \subseteq A_0\). Pick any point \(y'\) in \((c, e)\). Because \([a, c]\) can be covered by finitely many open sets, adding \(A_0\) yields a finite cover of the larger interval \([a, y']\). Thus \(y' \in S\). But this contradicts the fact that \(c\) is the supremum of \(S\) (since \(y' > c\)).
Therefore our assumption is false, and \(c = b\).
Combining Claim 1 and Claim 2, we obtain \(b \in S\), i.e., the whole closed interval \([a, b]\) can be covered by finitely many open sets from \(\mathcal{A}\). This completes the proof. \(\square\)
With this lemma, we can prove the most important compactness theorem in Euclidean spaces.
Theorem 9.2 (Heine–Borel Theorem)
In Euclidean space \(\mathbb{R}^n\) with the standard topology, a subset \(S \subseteq \mathbb{R}^n\) is compact if and only if it is closed and bounded.
Proof of Theorem 9.2 (Click to expand)
Sufficiency (closed and bounded \(\implies\) compact):
Suppose \(S\) is bounded. Then it can be enclosed in a sufficiently large \(n\)-dimensional closed cube, i.e., there exists a constant \(M > 0\) such that
By Lemma 9.1, the one-dimensional closed interval \([-M, M]\) is compact. From Chapter 8, the product of finitely many compact spaces (or by Tychonoff's theorem) \([-M, M]^n\) is again compact.
Since \(S\) is closed and is a closed subset of the compact space \([-M, M]^n\), by the property that closed subsets of compact spaces are compact (Theorem 8.1), \(S\) must be compact.
Necessity (compact \(\implies\) closed and bounded):
First, \(\mathbb{R}^n\) is a Hausdorff space. By Theorem 8.2 (compact subsets of Hausdorff spaces are closed), a compact subset \(S\) must be closed.
Next, we show \(S\) is bounded. Consider the following specific open cover of \(\mathbb{R}^n\):
Clearly this family of expanding open cubes covers all of \(\mathbb{R}^n\), and hence covers \(S\).
Because \(S\) is compact, this open cover has a finite subcover. Since the cubes are nested (\(N_1 \subset N_2 \subset \dots\)), the largest among these finitely many open sets contains \(S\).
Thus there exists a positive integer \(K\) such that
This directly shows \(S\) is bounded. \(\square\)
2. Compactness and Extreme Values of Continuous Functions
In calculus, a continuous function on a closed interval attains its maximum and minimum. Using the language of compactness, we can generalise this to any compact topological space.
Theorem 9.3 (Extreme Value Theorem)
Let \(X\) be a compact space and \(f: X \to \mathbb{R}\) a continuous real‑valued map. Then \(f\) attains both a maximum and a minimum on \(X\). That is, there exist \(c, d \in X\) such that for every \(x \in X\),
Proof of Theorem 9.3 (Click to expand)
Because \(f\) is continuous and \(X\) is compact, by Theorem 8.3 (continuous maps preserve compactness), the image \(f(X)\) is a compact subset of \(\mathbb{R}\).
By the Heine–Borel theorem (Theorem 9.2), a compact set in \(\mathbb{R}\) is closed and bounded.
- Since \(f(X)\) is bounded, by the completeness of \(\mathbb{R}\), it has a supremum \(\sup f(X)\) and an infimum \(\inf f(X)\).
- Since \(f(X)\) is closed, it contains all its limit points, hence it contains its supremum and infimum.
Thus there exists \(d \in X\) with \(f(d) = \sup f(X)\), and there exists \(c \in X\) with \(f(c) = \inf f(X)\).
Therefore for every \(x \in X\), we have \(f(c) \le f(x) \le f(d)\). The extreme values are attained. \(\square\)
3. Isolated Points and the Cardinality of Compact Hausdorff Spaces
When examining the point‑set structure of a topological space, isolated points are an important concept.
Definition 9.1 (Isolated Point)
In a topological space \((X, \tau)\), a point \(x \in X\) is called an isolated point if the singleton set \(\{x\}\) itself is open (i.e., \(\{x\} \in \tau\)).
A profound topological conclusion is that if a compact Hausdorff space has no isolated points (such spaces are sometimes called “perfect” compact spaces), then it must contain uncountably many points.
Theorem 9.4 (A compact Hausdorff space without isolated points is uncountable)
Let \((X, \tau)\) be a non‑empty compact Hausdorff space. If \(X\) has no isolated points, then \(X\) is uncountable.
Proof of Theorem 9.4 (Click to expand)
We first prove a key claim.
Claim: For any \(x \in X\) and any open set \(U\) containing \(x\), there exists a non‑empty open set \(V\) such that
Proof of the claim: Since \(X\) has no isolated points, the open set \(U\) cannot consist only of \(x\) itself. Hence there exists another point \(y \in U \setminus \{x\}\). Because \(X\) is Hausdorff, the distinct points \(x\) and \(y\) can be separated by disjoint open sets: there exist open sets \(W_1\) and \(W_2\) with
Define \(V = W_2 \cap U\). Clearly \(y \in V\), so \(V\) is a non‑empty open set. Since \(V \subseteq W_2\), its closure \(\bar{V} \subseteq \bar{W}_2\). Because \(x \in W_1\) and \(W_1\) is open and disjoint from \(W_2\), we have \(x \notin \bar{W}_2\), and consequently \(x \notin \bar{V}\). Moreover, since \(X\) is normal (every compact Hausdorff space is normal, hence regular), we can shrink \(V\) appropriately so that its closure is completely contained in \(U\). This proves the claim.
Now we prove the main theorem by contradiction.
Assume \(X\) is countable. Since \(X\) is non‑empty, we can list all points of \(X\) as a sequence:
Using the claim, we construct an infinite nested sequence of open sets.
-
Step 1: Take \(U_1 = X\). By the claim, for the point \(x_1\) and the open set \(U_1\), there exists a non‑empty open set \(V_1\) such that \(\bar{V}_1 \subseteq U_1\) and \(x_1 \notin \bar{V}_1\).
-
Step 2: Now consider the non‑empty open set \(V_1\) and the next point \(x_2\). If \(x_2 \notin V_1\), we arbitrarily choose a non‑empty open set inside \(V_1\) whose closure is contained in \(V_1\) and call it \(V_2\). If \(x_2 \in V_1\), then by the claim there exists a non‑empty open set \(V_2\) such that \(\bar{V}_2 \subseteq V_1\) and \(x_2 \notin \bar{V}_2\).
-
Inductively: Suppose we have constructed a non‑empty open set \(V_{n-1}\). For the point \(x_n\), we can always find a non‑empty open set \(V_n\) satisfying
Thus we obtain a descending chain of closed sets:
Since \(X\) is compact and these closures are non‑empty closed sets, they satisfy the finite intersection property (FIP). In a compact space, the global intersection of any family of closed sets with the FIP must be non‑empty!
Hence there exists a point \(y\) such that
Because \(y \in X\) and we assumed \(X = \{x_1, x_2, \dots\}\), \(y\) must be some point of the sequence, say \(y = x_k\).
But this leads to a contradiction! In the \(k\)‑th step of our construction, we explicitly required
Therefore \(x_k\) cannot belong to the intersection of all the closures.
This contradiction refutes the assumption that \(X\) is countable. Hence a compact Hausdorff space without isolated points must be uncountable. \(\square\)
4. Nested Closed Set Theorem
At the end of the proof of the previous theorem, we used an important corollary about closed sets in compact spaces. We state it here explicitly as a summary of this chapter.
Corollary 9.5 (Nested Closed Set Theorem in Compact Spaces)
In a compact topological space \((X, \tau)\), let \(\{C_n\}_{n=1}^\infty\) be a descending chain of non‑empty closed sets, i.e.,
Then the infinite intersection of this chain is non‑empty:
This corollary is essentially a direct application of the finite intersection property (FIP). It plays a central role not only in point‑set topology but also bears a striking resemblance to the “nested ball theorem” in complete metric spaces studied in functional analysis. The key difference is that the nested ball theorem in functional analysis relies on the completeness of the metric and the radii tending to zero, whereas the nested closed set theorem in topology depends purely on compactness.