What are Compact and Precompact in Topological Spaces?
Definition
Let a topological space be $\left( X, \mathscr{T} \right)$, and denote $A \subset X$.
- A collection comprised of open sets of $X$ is called an open covering of $A$ if it satisfies the following condition: $$ A \subset \bigcup_{O \in \mathscr{O}} O $$
- A subset $\mathscr{O} ' $ of $\mathscr{O}$ is called a subcover, especially when the cardinality of $\mathscr{O} ' $ is a natural number, it’s termed a finite subcover.
- If every open cover of $X$ has a finite subcover, then $X$ is said to be compact. In other words, if there exists a finite set $\mathscr{O} ' = \left\{ O_{1} , \cdots , O_{n} \right\} \subset \mathscr{O}$ satisfying the following for every open cover $\mathscr{O}$, then $X$ is compact. $$ X = \bigcup_{i=1}^{n} O_{i} $$
- If $A$, as a subspace of $X$, is compact, then $A$ is called compact.
- Let $X$ be a topological space. If the closure $\overline{K}$ of a subset $K \subset X$ is compact, then $K$ is said to be precompact or relatively compact.
Explanation
Compact
Given the utility of the condition compact in analysis, it is only natural to pursue its generalization. Although the language has become more complex with generalization, the essential part remains unchanged.
In fact, compactness is crucial in various theories. The fact that a set is compact signifies that it can be thought of as divided into finite pieces, making it a favorable condition for rigorous proofs. Conversely, when proving a theorem, showing that a set $A$ is indeed compact often becomes a key issue.
Precompact
Precompact is a term that aptly describes a set $K$ that may not be compact by itself but becomes compact when its closure is taken. In metric spaces, it is also referred to as a totally bounded space, while the term relatively compact stems from the relative nature of closure itself. If $K$ becomes the entire space rather than a subspace of $X$, then $K$ is closed in itself, so $K = \overline{K}$, and hence the statement that $\overline{K}$ is compact means that $K$ is (relatively) compact.
Alternatively, precompact can be defined using sequences. The definition is as follows:
That $K \subset X$ is precompact means that every sequence $\left\{ x_{n} \right\} \subset K$ defined in $K$ has a subsequence $\left\{ x_{n '} \right\} \subset \left\{ x_{n} \right\}$ converging to $x \in X$.
Expressing this in formulae is as follows:
$$ K : \text{precompact} \iff \forall \left\{ x_{n} \right\} \subset K, \exists \left\{ x_{n '} \right\} \subset \left\{ x_{n} \right\} : x_{n '} \to x \in X \text{ as } n \to \infty $$
Particularly, when the condition specifies $x \in K$ rather than $x \in X$, $K$ is called sequentially compact.
Theorems
- [1]: That $A$ is compact is equivalent to the fact that every open cover of $A$ has a finite subcover.
- [2]: If a subset $F$ of a compact set $K$ is closed, then $F$ is compact.
- [3]: That $X$ is compact is equivalent to the fact that when every family of sets containing only closed sets of $X$ has a finite intersection property, the intersection itself is non-empty.
Proofs
[1]
$\Gamma$ is an index set.
$( \implies )$
Assume $A \subset X$ is compact and $\mathscr{U} := \left\{ U_{\alpha} : \alpha \in \Gamma \right\}$ is an open cover of $A$. Then $U_{\alpha} \cap A$ is an open set in the subspace $A$ of $X$, and $\mathscr{O} := \left\{ U_{\alpha} \cap A : U_{\alpha} \in \mathscr{U} \right\}$ becomes an open cover of $A$. Since $A$ is compact, there exists $\alpha_{1} , \cdots , \alpha_{n} \in \Gamma$ that satisfies $\displaystyle A \subset \bigcup_{i=1}^{n} \left( U_{\alpha_{i}} \cap A \right)$. Then, it can be confirmed that $\left\{ U_{\alpha_{1}} , \cdots , U_{\alpha_{n}} \right\}$ exists as a finite subcover of $\mathscr{U}$.
$( \impliedby )$
Consider an open cover $\mathscr{O} := \left\{ O_{\alpha} : O_{\alpha} \text{ is open in } A, \alpha \in \Gamma \right\}$ consisting of open sets in $A$. Since $O_{\alpha}$ is open in $A$, for each $\alpha \in \Gamma$, there exists an open set $U_{\alpha}$ satisfying $U_{\alpha} \cap A = O_{\alpha}$. The set of these $\mathscr{U} := \left\{ U_{\alpha} : \alpha \in \Gamma \right\}$ forms an open cover of $A$. According to the assumption, every open cover has a finite subcover $\left\{ U_{\alpha_{1}} , \cdots , U_{\alpha_{n}} \right\}$, so $\left\{ O_{\alpha_{1}} , \cdots , O_{\alpha_{n}} \right\}$ becomes a finite subcover of $\mathscr{O}$.
■
[2]
$$ F \subset K \subset X $$ Suppose $F$ is a closed set in $X$ and has an open cover $\left\{ U_{\alpha} \right\}_{\alpha}$, and $K$ is compact. Since $F$ is closed, $F^{c}$ is open in $X$, thus $\left\{ F^{c} \right\} \cup \left\{ U_{\alpha} \right\}_{\alpha}$ becomes one of the open covers of $K$. Since $K$ is compact, there exists a finite subcover $\Phi$ of $\left\{ F^{c} \right\} \cup \left\{ U_{\alpha} \right\}_{\alpha}$ that satisfies $F \subset K \subset \Phi$.
- If $F^{c}\notin \Phi$, then $\Phi$ is a finite subcover of $\left\{ U_{\alpha} \right\}_{\alpha}$, making $F$ compact.
- If $F^{c}\in \Phi$, then $\Phi \setminus \left\{ F^{c} \right\}$ becomes a finite subcover of $\left\{ U_{\alpha} \right\}_{\alpha}$, making $F$ compact.
Since $F$ is compact in both possible scenarios, $F$ is compact.
■
[3]
Strategy: Understanding the wording is the key due to its complexity. The fact that $\mathscr{C}$ has a finite intersection property does not guarantee $\displaystyle \bigcap_{C \in \mathscr{C}} C \ne \emptyset$ is needed, and the compact condition is necessary. Note that the definition of compact deals with the union of open sets, while this theorem deals with the intersection of closed sets. From these considerations, one can grasp how the finite intersection property might relate to compactness and proceed with the proof.
$\Gamma$ is an index set.
Finite Intersection Property: That a family of sets $\mathscr{A} \subset \mathscr{P}(X)$ consisting of subsets of $X$ has a finite intersection property (f.i.p, finite intersection property) means that every finite subset $A \subset \mathscr{A}$ of $\mathscr{A}$ has a non-empty intersection. Mathematically, it is expressed as: $$ \forall A \subset \mathscr{A}, \bigcap_{a \in A} a \ne \emptyset $$
$( \implies )$
Assume $X$ is compact and $\mathscr{C} := \left\{ C_{\alpha} : C_{\alpha} \text{ is closed in } X, \alpha \in \Gamma \right\}$ has a finite intersection property. Now, assume $\displaystyle \bigcap_{\alpha \in \Gamma} C_{\alpha} = \emptyset$ and set $\mathscr{O} := \left\{ X \setminus C_{\alpha} : C_{\alpha} \in \mathscr{C} \right\}$. Then $$ \begin{align*} \bigcup_{\alpha \in \Gamma} ( X \setminus C_{\alpha}) =& X \setminus \bigcap_{\alpha \in \Gamma} C_{\alpha} \\ =& X \setminus \emptyset \\ =& X \end{align*} $$ making $\mathscr{O}$ an open cover of $X$. Since $X$ is compact, $\mathscr{O}$ has a finite subcover $\displaystyle \left\{ (X \setminus C_{\alpha_{1}}) , \cdots ,(X \setminus C_{\alpha_{n}}) \right\}$, which implies that $$ X = \bigcup_{i=1}^{n} ( X \setminus C_{\alpha_{i}}) = X \setminus \bigcap_{i=1}^{n} C_{\alpha_{i}} $$ and thus contradicts the assumption that $\mathscr{C}$ has a finite intersection property. Therefore, $\displaystyle \bigcap_{C \in \mathscr{C}} C \ne \emptyset$ must hold.
$( \impliedby )$
Consider an open cover $\mathscr{O} := \left\{ O_{\alpha} : O_{\alpha} \text{ is open in } A, \alpha \in \Gamma \right\}$ of $X$ and $\mathscr{C} := \left\{ X \setminus O_{\alpha} : O_{\alpha} \in \mathscr{O} \right\}$. $$ \begin{align*} \bigcap_{\alpha \in \Gamma} C_{\alpha} &= \bigcap_{\alpha \in \Gamma} ( X \setminus O_{\alpha}) \\ =& X \setminus \bigcup_{\alpha \in \Gamma} O_{\alpha} \\ =& X \setminus X \\ =& \emptyset \end{align*} $$ Thus, by contrapositive law, $\mathscr{C}$ does not have a finite intersection property. This implies there exists $C_{\alpha_{1}} , \cdots , C_{\alpha_{n}} \in \mathscr{C}$ satisfying $\displaystyle \bigcap_{i=1}^{n} C_{\alpha_{i}} = \emptyset$. Then $$ \begin{align*} X \setminus \bigcup_{i=1}^{n} O_{i} =& X \setminus \bigcup_{i=1}^{n} (X \setminus C_{i}) \\ =& X \setminus \left( X \setminus \bigcap_{i=1}^{n} C_{i} \right) \\ =& \bigcap_{i=1}^{n} C_{i} \\ =& \emptyset \end{align*} $$ so $\displaystyle X = \bigcup_{i=1}^{n} O_{i}$. In other words, since there exists a finite subcover for the open cover $\mathscr{O}$, it is compact.
■