Finite Intersection Property Criterion for Compactness in a Topo. Sp.

Finite Intersection Property Criterion for Compactness in a Topological Space

Recall from the Compactness of Sets in a Topological Space page that if $X$ is a topological space and $A \subseteq X$ then $A$ is said to be compact in $X$ if every open cover of $A$ has a finite subcover.

We will now look at a nice criterion for a set $A$ to be compact in a topological space $X$. We will first define a special property of a collection of sets as it will be used in the criterion theorem.

Definition: Let $\mathcal F$ be a collection of sets. Then $\mathcal F$ is said to have the Finite Intersection Property if for every finite collection of sets from $\mathcal F$, $\{ F_1, F_2, ..., F_n \} \subseteq \mathcal F$, we have that $\displaystyle{\bigcap_{i=1}^{n} F_i \neq \emptyset}$.
Theorem 1: Let $X$ be a topological space. Then $X$ is compact if and only if for every collection of closed sets $\mathcal F$ from $X$, we have that if $\mathcal F$ has the finite intersection property then $\displaystyle{\bigcap_{F \in \mathcal F} F \neq \emptyset}$.
  • Proof: $\Rightarrow$ Suppose that $X$ is compact. Let $\mathcal F$ be a collection of closed sets from $X$. Suppose that $\mathcal F$ has the finite intersection property. Then for every finite subcollection $\{ F_1, F_2, ..., F_n \} \subseteq \mathcal F$ we have that:
(1)
\begin{align} \quad \bigcap_{i=1}^{n} F_i \neq \emptyset \end{align}
  • We want to show that $\displaystyle{\bigcap_{F \in \mathcal F} F \neq \emptyset}$. Suppose not, i.e., suppose that:
(2)
\begin{align} \quad \bigcap_{F \in \mathcal F} F = \emptyset \end{align}
  • Then taking the complement of both sides above and using De Morgan's laws and we see that:
(3)
\begin{align} \quad \left ( \bigcap_{F \in \mathcal F} F \right )^c & = \emptyset^c \\ \quad \bigcup_{F \in \mathcal F} F^c & = X \end{align}
  • Since $F$ is closed for all $F \in \mathcal F$ we have that $F^c$ is open for all $F \in \mathcal F$. So $\{ F^c : F \in \mathcal F \}$ is an open cover of $X$. Since $X$ is compact, there exists a finite subcover, say $\{ F_1^c, F_2^c, ..., F_n^c \} \subseteq \{ F^c : F \in \mathcal F \}$ such that:
(4)
\begin{align} \quad \bigcup_{i=1}^{n} F_i^c = X \end{align}
  • Taking the complement of both sides again and using De Morgan's laws and we have that:
(5)
\begin{align} \quad \left ( \bigcup_{i=1}^{n} F_i^c \right )^c & = X^c \\ \quad \bigcap_{i=1}^{n} F_i & = \emptyset \end{align}
  • But then $\{ F_1, F_2, ..., F_n \}$ is a finite collection of sets from $\mathcal F$ that does not have the finite intersection property which is a contradiction. Hence the assumption that $\displaystyle{\bigcap_{F \in \mathcal F} F = \emptyset}$ was false. So $\displaystyle{\bigcap_{F \in \mathcal F} F \neq \emptyset}$.
  • $\Leftarrow$ Suppose that if for every collection $\mathcal F$ of closed sets from $X$ we have that if $\mathcal F$ has the finite intersection property then $\displaystyle{\bigcap_{F \in \mathcal F} F \neq \emptyset}$.
  • Let $\{ A_i : i \in I \}$ be an open cover of $X$. Then:
(6)
\begin{align} \quad \bigcup_{i \in I} A_i = X \end{align}
  • Suppose $X$ is not compact. Then for all finite subsets $J \subseteq I$ we have that:
(7)
\begin{align} \quad \bigcup_{i \in J} A_i \neq X \end{align}
  • Taking the complement of both sides of the equation above and using De Morgan's laws shows us that:
(8)
\begin{align} \quad \left ( \bigcup_{i \in J} A_i \right )^c & \neq X^c \\ \quad \bigcap_{i \in J} A_i^c & \neq \emptyset \end{align}
  • So $\{ A_i^c : i \in I \}$ has the finite intersection property. Therefore:
(9)
\begin{align} \quad \bigcap_{i \in I} A_i^c \neq \emptyset \end{align}
  • Taking the complement of both sides of the equation above and using De Morgan's laws gives:
(10)
\begin{align} \quad \left ( \bigcap_{i \in I} A_i^c \right )^c & \neq \emptyset^c \\ \quad \bigcup_{i \in I} A_i & \neq X \end{align}
  • But this contradicts $\{ A_i : i \in I \}$ being an open cover of $X$. Hence the assumption that $X$ is not compact was false. Therefore $X$ is compact. $\blacksquare$
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License