What are Compact and Precompact in Topological Spaces?
Definition 1
Let’s say we have a topological space $\left( X, \mathscr{T} \right)$.
- A set $\mathscr{O} \subset \mathscr{T}$ consisting of open sets of $X$ is called an open covering of $A$ if it satisfies the following: $$ A \subset \bigcup_{O \in \mathscr{O}} O $$
- A subset $\mathscr{O} ' $ of $\mathscr{O}$ is called a subcover of $\mathscr{O}$. If the cardinality of $\mathscr{O} ' $ is a natural number, it is called a finite subcover.
- $X$ is said to be compact if every open cover of $X$ has a finite subcover. In other words, for every open cover $\mathscr{O}$, if there exists a finite set $\mathscr{O} ' = \left\{ O_{1} , \cdots , O_{n} \right\} \subset \mathscr{O}$ satisfying the following, then $X$ is compact: $$ X = \bigcup_{i=1}^{n} O_{i} $$
- If a subspace $A$ of $X$ is compact as a subspace, then $A$ is said to be compact.
- Let’s consider $X$ as a topological space. A subset $K$ is said to be precompact or relatively compact if the closure $\overline{K}$ of $K$ is compact.
Explanation
Compactness
Considering how useful the condition of compactness was in introductory analysis, it’s natural to seek its generalization. While the generalization might have made the terminology more complex, the essence hasn’t changed.
In fact, compactness is crucially applied in various theories. Being compact implies that a set can be considered in finite pieces, making it an excellent condition for rigorous proofs. Conversely, demonstrating that a certain set $A$ appearing in a proof is indeed compact often becomes a key part of the argument.
Precompactness
Precompactness suggests that even if $K$ itself is not compact, taking the closure of $K$ results in a compact set, indicating the potential for $K$ to become compact. In metric spaces, this is also known as a totally bounded space. The alternative term relatively compact comes from the relative nature of closure; if $K$ is considered not as a subspace but as the entire space itself, then $K$ is closed in $K$, making $\overline{K}$ compact, which essentially means $K$ is (relatively) compact.
Precompactness can also be defined in terms of sequences:
$K \subset X$ is precompact if, for every sequence $\left\{ x_{n} \right\} \subset K$ defined in $K$, there exists a subsequence $\left\{ x_{n '} \right\} \subset \left\{ x_{n} \right\}$ that converges to $x \in X$.
Expressed mathematically as:
$$ K : \text{precompact} \iff \forall \left\{ x_{n} \right\} \subset K, \exists \left\{ x_{n '} \right\} \subset \left\{ x_{n} \right\} : x_{n '} \to x \in X \text{ as } n \to \infty $$
Especially, when the condition involves $x \in K$ instead of $x \in X$, $K$ is termed sequentially compact.
Theorem
- [1]: $A$ being compact is equivalent to every open cover of $A$ having a finite subcover.
- [2]: If a subset $F$ of a compact set $K$ is closed, then $F$ is compact.
- [3]: $X$ being compact is equivalent to every family of closed sets in $X$ having the finite intersection property (f.i.p.), meaning taking intersections over any collection of these sets results in a non-empty set.
Proof
[1]
$\Gamma$ represents 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 $\alpha_{1} , \cdots , \alpha_{n} \in \Gamma$ satisfying $\displaystyle A \subset \bigcup_{i=1}^{n} \left( U_{\alpha_{i}} \cap A \right)$. Hence, $\left\{ U_{\alpha_{1}} , \cdots , U_{\alpha_{n}} \right\}$ exists as a finite subcover of $\mathscr{U}$.
$( \impliedby )$
Consider an open cover $\mathscr{O} := \left\{ O_{\alpha} : O_{\alpha} \text{ is open in } A, \alpha \in \Gamma \right\}$ of $A$ consisting of open sets. For each $\alpha \in \Gamma$, there exists an open set $U_{\alpha}$ satisfying $U_{\alpha} \cap A = O_{\alpha}$. The set of these $U_{\alpha}$, denoted as $\mathscr{U} := \left\{ U_{\alpha} : \alpha \in \Gamma \right\}$, forms an open cover of $A$. Since every open cover has a finite subcover $\left\{ U_{\alpha_{1}} , \cdots , U_{\alpha_{n}} \right\}$, $\left\{ O_{\alpha_{1}} , \cdots , O_{\alpha_{n}} \right\}$ becomes a finite subcover of $\mathscr{O}$.
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[2]
$$ F \subset K \subset X $$ Assume $F$ is a closed set in $X$ and $K$ is compact. Consider an open cover $\left\{ U_{\alpha} \right\}$ of $F$. Since $F$ is closed, $F^{c}$ is an open set in $X$ and hence $F^{c} \cup \left\{ U_{\alpha} \right\}$ becomes one of the open covers of $K$. Since $K$ is compact, there exists a finite subcover $\Phi$ satisfying $F \subset K \subset \Phi$.
- If $F^{c}\notin \Phi$, then $\Phi$ becomes a finite subcover of $\left\{ U_{\alpha} \right\}$, making $F$ compact.
- If $F^{c}\in \Phi$, then $\Phi \setminus \left\{ F^{c} \right\}$ becomes a finite subcover of $\left\{ U_{\alpha} \right\}$, making $F$ compact.
In both possible cases, $F$ is compact.
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[3]
Strategy: The terminology here is quite complex, so understanding the terms is crucial. Having the finite intersection property (f.i.p.) for $\mathscr{C}$ does not guarantee $\displaystyle \bigcap_{C \in \mathscr{C}} C \ne \emptyset$; the compact condition is necessary. The definition of compactness involves unions of open sets, while this theorem involves intersections of closed sets, highlighting the relationship between finite intersection properties and compactness.
$\Gamma$ is an index set.
Finite Intersection Property: For a family $\mathscr{A} \subset \mathscr{P}(X)$ of subsets of $X$, having the finite intersection property (f.i.p) means that for every finite subfamily $A \subset \mathscr{A}$, the intersection of its members is non-empty. This is expressed mathematically as: $$ \forall A \subset \mathscr{A}, \bigcap_{a \in A} a \ne \emptyset $$
$( \implies )$
Assuming $X$ is compact and $\mathscr{C} := \left\{ C_{\alpha} : C_{\alpha} \text{ is closed in } X, \alpha \in \Gamma \right\}$ has the f.i.p., suppose $\displaystyle \bigcap_{\alpha \in \Gamma} C_{\alpha} = \emptyset$ and choose $\mathscr{O} := \left\{ X \setminus C_{\alpha} : C_{\alpha} \in \mathscr{C} \right\}$. 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 $$ X = \bigcup_{i=1}^{n} ( X \setminus C_{\alpha_{i}}) = X \setminus \bigcap_{i=1}^{n} C_{\alpha_{i}} $$, leading to $\displaystyle \bigcap_{i=1}^{n} C_{\alpha_{i}} = \emptyset$. This contradicts the assumption that $\mathscr{C}$ has the f.i.p., thus $\displaystyle \bigcap_{C \in \mathscr{C}} C \ne \emptyset$ must hold.
$( \impliedby )$
Consider the open cover $\mathscr{O} := \left\{ O_{\alpha} : O_{\alpha} \text{ is open in } A, \alpha \in \Gamma \right\}$ of $X$ and $\mathscr{C} := \left\{ X \setminus O_{\alpha} : O_{\alpha} \in \mathscr{O} \right\}$. Since $$ \begin{align*} \bigcap_{\alpha \in \Gamma} C_{\alpha} &= \bigcap_{\alpha \in \Gamma} ( X \setminus O_{\alpha}) \\ =& X \setminus \bigcup_{\alpha \in \Gamma} O_{\alpha} \\ =& X \setminus X \\ =& \emptyset \end{align*} $$, by contraposition, $\mathscr{C}$ lacks the f.i.p., meaning there exists $C_{\alpha_{1}} , \cdots , C_{\alpha_{n}} \in \mathscr{C}$ satisfying $\displaystyle \bigcap_{i=1}^{n} C_{\alpha_{i}} = \emptyset$. Then, $$ \begin{align*} X \setminus \bigcup_{i=1}^{n} O_{i} =& X \setminus \bigcup_{i=1}^{n} (X \setminus C_{i}) \\ =& X \setminus \left( X \setminus \bigcap_{i=1}^{n} C_{i} \right) \\ =& \bigcap_{i=1}^{n} C_{i} \\ =& \emptyset \end{align*} $$, making $\displaystyle X = \bigcup_{i=1}^{n} O_{i}$. Hence, having a finite subcover for every open cover implies compactness.
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See Also
Munkres. (2000). Topology (2nd Edition): p164. ↩︎