Chapter 8: Compactness
In point-set topology, compactness is the perfect generalization of the concept of "bounded closed intervals" in calculus to abstract topological spaces. Compact spaces possess extremely excellent properties; they not only guarantee that continuous real-valued functions achieve maximum and minimum values but also impose strict restrictions on the products and mappings of topological spaces. This chapter will explore the definition of compactness in detail, its profound connection with closed sets and Hausdorff spaces, and its preservation under continuous mappings and product spaces.
1. Basic Concepts of Compactness
In calculus, a continuous function on a closed interval \([a, b]\) must be bounded and able to reach its extreme values. To generalize this property to general topological spaces, we introduce the concepts of open covers and finite subcovers.
Definition 8.1 (Open Cover and Compactness)
- Open Cover: Let \(X\) be a topological space, and \(\mathcal{A} = \{A_i\}_{i \in I}\) be a family of open sets. If the union of these open sets contains the entire space \(X\), namely:
then \(\mathcal{A}\) is called an open cover of \(X\).
- Compact Space: If for any open cover \(\mathcal{A}\) of a topological space \(X\), there necessarily exists a finite subcover, i.e., there exist finitely many indices \(i_1, i_2, \dots, i_k \in I\) such that:
then \(X\) is called a compact space.
- Compact Subset: If a subset \(Y\) of \(X\) is compact as a subspace (endowed with the relative topology), then \(Y\) is called a compact subset of \(X\). This is equivalent to saying that for any family of open sets in \(X\) that covers \(Y\), a finite number of open sets can always be selected from them to still cover \(Y\).
Classic Counterexample: The real line \(\mathbb{R}\) (endowed with the standard topology) is not a compact space. Consider the open cover \(\mathcal{A} = \{(-n, n) \mid n \in \mathbb{N}_{\ge 1}\}\). Obviously, this family of open intervals covers the entire \(\mathbb{R}\). However, if we take any finite number of intervals from it, the largest one will necessarily be of the form \((-N, N)\), which clearly cannot cover the entire real line. Therefore, \(\mathbb{R}\) is not compact.
2. Relationship between Compact Sets and Closed Sets
There is an extremely close connection between compactness and closed sets, especially when the space possesses certain separation properties (Hausdorff property).
Theorem 8.1 (Closed Sets in Compact Spaces)
Let \(X\) be a compact space. If \(Y \subseteq X\) is a closed set, then \(Y\) is also compact.
Proof of Theorem 8.1
Let \(\mathcal{U} = \{U_i\}_{i \in I}\) be a family of open sets in \(X\), and they cover the subset \(Y\), namely \(Y \subseteq \bigcup_{i \in I} U_i\).
Because \(Y\) is a closed set in \(X\), its complement \(X \setminus Y\) is an open set in \(X\).
We add \(X \setminus Y\) to the original family of covers to construct a new family of open sets \(\mathcal{U}' = \{U_i\}_{i \in I} \cup \{X \setminus Y\}\). Obviously, \(\mathcal{U}'\) covers the entire space \(X\).
Since \(X\) is a compact space, this open cover \(\mathcal{U}'\) must have a finite subcover.
In this finite subcover, removing \(X \setminus Y\) if it exists leaves what must be a finite number of open sets from the original cover \(\mathcal{U}\), and they still cover \(Y\).
Therefore, any open cover of \(Y\) has a finite subcover, so \(Y\) must be compact. \(\square\)
In general topological spaces, compact sets are not necessarily closed (for example, in a trivial topology). However, if we add the Hausdorff separation axiom, compact sets are necessarily closed.
Theorem 8.2 (Compact Sets in Hausdorff Spaces)
Let \(X\) be a Hausdorff space. If \(Y \subseteq X\) is a compact subset, then \(Y\) must be a closed set in \(X\).
Proof of Theorem 8.2
We need to prove that the complement \(X \setminus Y\) of \(Y\) is an open set. Take any point \(x \in X \setminus Y\).
Because \(X\) is a Hausdorff space, for any point \(y \in Y\) in \(Y\), point \(x\) and point \(y\) are two distinct points; therefore, disjoint open neighborhoods exist: there exists an open set \(U_y\) containing \(x\) and an open set \(V_y\) containing \(y\) such that:
When we let \(y\) traverse the entire \(Y\), the family of sets \(\{V_y \mid y \in Y\}\) clearly constitutes an open cover of \(Y\).
Because \(Y\) is a compact set, there exist finitely many points \(y_1, y_2, \dots, y_k \in Y\) such that the corresponding finite number of open sets cover \(Y\):
For these finitely many points, they correspond respectively to the finite number of open sets \(U_{y_1}, U_{y_2}, \dots, U_{y_k}\) containing \(x\). We take their intersection:
Since this is an intersection of finitely many open sets, \(U\) is still an open set containing \(x\).
More importantly, because \(U_{y_i} \cap V_{y_i} = \emptyset\) for every \(i\), the intersection \(U\) does not intersect any of the \(V_{y_i}\). Thus \(U\) does not intersect the union of the finite cover, and naturally does not intersect \(Y\):
This means \(x \in U \subseteq X \setminus Y\). Because an open neighborhood \(U\) completely contained in the complement can be found for any point \(x\) in the complement, \(X \setminus Y\) is an open set, and thus \(Y\) is a closed set. \(\square\)
3. Compactness and Continuous Mappings
The core of topology lies in studying how continuous mappings preserve the structure of a space. Just as continuous mappings can maintain connectedness, they also perfectly preserve compactness.
Theorem 8.3 (Continuous Mappings Preserve Compactness)
Let \(f: X \to Y\) be a continuous mapping between topological spaces. If the domain \(X\) is compact, then the image set \(f(X)\) is also compact in \(Y\).
Proof of Theorem 8.3
Let \(\mathcal{V} = \{V_i\}_{i \in I}\) be an open cover of the image set \(f(X)\) in \(Y\) (\(V_i\) are open sets in \(Y\)).
Since \(f\) is a continuous mapping, the preimage \(f^{-1}(V_i)\) of the open set \(V_i\) under \(f\) must be an open set in the domain \(X\).
Since \(\{V_i\}_{i \in I}\) covers \(f(X)\), their family of preimages \(\{f^{-1}(V_i)\}_{i \in I}\) constitutes an open cover of the entire space \(X\), namely:
\(X\) is known to be a compact space; therefore, this open cover must have a finite subcover. There exist finitely many indices \(i_1, i_2, \dots, i_k \in I\) such that:
Applying the mapping \(f\) to both sides of the equation simultaneously, we obtain:
This finds a finite subcover of the image set \(f(X)\). Therefore, \(f(X)\) is compact. \(\square\)
Generalization of the Extreme Value Theorem: If \(X\) is a compact space and there exists a continuous real-valued function \(f: X \to \mathbb{R}\), it is known from Theorem 8.3 that \(f(X)\) is a compact set in \(\mathbb{R}\). In \(\mathbb{R}\), compact sets are equivalent to bounded closed sets (Heine-Borel Theorem). Since \(f(X)\) is bounded and closed, it must contain its supremum and infimum; that is, the function \(f\) must be able to achieve a maximum value and a minimum value on the compact space \(X\).
Combining Theorems 8.1, 8.2, and 8.3, we can derive a corollary that is extremely important in the classification of topological spaces:
Theorem 8.4 (Continuous Bijection from a Compact Space to a Hausdorff Space is a Homeomorphism)
Let \(X\) be a compact space and \(Y\) be a Hausdorff space. If \(f: X \to Y\) is a continuous bijection, then \(f\) must be a homeomorphism.
Proof of Theorem 8.4
\(f\) is known to be bijective and continuous. To prove that \(f\) is a homeomorphism, it only needs to be proven that the inverse mapping \(f^{-1}: Y \to X\) is also continuous.
This is equivalent to proving: for any closed set \(A\) in \(X\), its "preimage" under \(f^{-1}\)βwhich is \(f(A)\)βmust be a closed set in \(Y\). This is referred to as mapping \(f\) being a "closed map."
-
Let \(A \subseteq X\) be any closed set.
-
Because \(X\) is a compact space, according to Theorem 8.1, the closed subset \(A\) is also compact.
-
Because \(f\) is continuous, according to Theorem 8.3, the continuous image \(f(A)\) of the compact set \(A\) is also compact.
-
Because the codomain \(Y\) is a Hausdorff space, according to Theorem 8.2, the compact subset \(f(A)\) must be a closed set in \(Y\).
Since \(f(A)\) is a closed set for any closed set \(A\), the inverse mapping \(f^{-1}\) is continuous. Therefore, \(f\) is a homeomorphism. \(\square\)
4. Compactness of Product Spaces and the Tube Lemma
The finite product of two compact spaces remains compact. This is intuitively obvious (for example, the product of two bounded closed intervals is a bounded closed rectangle), but in a general topological proof, we need to rely on a powerful tool: the "Tube Lemma."
εΌη 8.1 (Tube Lemma)
Let \(X\) be any topological space and \(Y\) be a compact space. If an open set \(N\) exists in the product space \(X \times Y\) that contains an entire "vertical slice" \(\{x_0\} \times Y\) generated by point \(x_0 \in X\), namely:
then there must exist an open neighborhood \(W\) of \(x_0\) in \(X\) such that the corresponding entire "tubular region" \(W \times Y\) is completely contained within \(N\), namely:
With the Tube Lemma, we can naturally prove the compactness of the product.
Theorem 8.5 (Compactness of Finite Products)
Let \(X\) and \(Y\) both be compact spaces; then the Cartesian product \(X \times Y\) endowed with the product topology must also be a compact space.
Proof of Theorem 8.5
Let \(\mathcal{A} = \{A_i\}_{i \in I}\) be any open cover of the product space \(X \times Y\).
First, fix any point \(x \in X\) in the domain \(X\). Examine the "slice" \(\{x\} \times Y\).
Since \(\{x\} \times Y\) is homeomorphic to the compact space \(Y\), \(\{x\} \times Y\) is also compact.
Since the slice is compact and \(\mathcal{A}\) covers the entire space and naturally also covers the slice, there must exist finitely many open sets \(A_{i_1}, A_{i_2}, \dots, A_{i_m} \in \mathcal{A}\) covering the slice \(\{x\} \times Y\).
Let the union of these finitely many open sets be \(N_x = \bigcup_{k=1}^m A_{i_k}\). Obviously, \(N_x\) is an open set containing the slice \(\{x\} \times Y\).
At this point, apply the Tube Lemma: there exists an open neighborhood \(W_x\) of \(x\) in \(X\) such that:
This indicates that for every \(x \in X\), an open neighborhood \(W_x\) containing \(x\) exists such that the entire "tubular region" \(W_x \times Y\) can be covered by a finite number of open sets from the original open cover \(\mathcal{A}\).
Now let \(x\) traverse the entire \(X\); the family of sets \(\{W_x \mid x \in X\}\) clearly constitutes an open cover of \(X\).
Utilizing the prerequisite condition: \(X\) is also a compact space. Therefore, finitely many points \(x_1, x_2, \dots, x_p \in X\) exist such that the corresponding finite number of open neighborhoods cover \(X\):
Then the entire product space can be written as the union of these finitely many "tubular regions":
Because each \(W_{x_j} \times Y\) can be covered by a finite number of open sets in \(\mathcal{A}\), the total union of these finitely many "tubular regions" \(X \times Y\) can also be covered by a finite number of open sets in \(\mathcal{A}\).
This proves that any open cover of \(X \times Y\) has a finite subcover; thus \(X \times Y\) is a compact space. \(\square\)
(Note: Through mathematical induction, the product of any finite number of compact spaces is compact. The conclusion that generalizes this to the product of an infinite number of compact spaces remaining compactβthe famous Tychonoff Theoremβwill rely on more complex set-theoretic tools, such as the Axiom of Choice and the Finite Intersection Property.)