📂Topology

Finite Intersection Property

Definition ¹

For a topological space $X$, let $\mathscr{A} \subset \mathscr{P} (X)$. If $\displaystyle \bigcap A \ne \emptyset$ for every finite subset $A \subset \mathscr{A}$, then $A$ is said to have the finite Intersection Property.

Explanation

That $A$ has the f.i.p. is equivalent to saying that for open sets $U_{\alpha} \subset A$ the following always holds. $$ \bigcap_{i=1}^{n} \left( X \setminus U_{i} \right) \ne \emptyset \implies \bigcap_{\alpha \in \forall } \left( X \setminus U_{\alpha} \right) \ne \emptyset $$

Note that this is a property of a set in general, not of a topological space. For example, $\displaystyle \left\{ \left. \left[ 0 , {{1} \over {n}} \right] \ \right| \ n \in \mathbb{N} \right\}$ can be said to have the f.i.p. without being endowed with any topology whatsoever.

The following theorem is useful because it gives an equivalent condition for compactness, but it is verbose even to state, and its proof is also very difficult to understand. Even if it makes you feel demoralized, compactness is inherently difficult, so just take it as it is.

Theorem

$X$ is compact if and only if it has the f.i.p. and $\displaystyle \bigcap_{\alpha \in \forall} A_{\alpha} \ne \emptyset$ holds for every closed $A_{\alpha} \subset X$.

Proof

For open sets $O_{\alpha}$ and closed sets $C_{\alpha}$, the following holds. $$ X \setminus \left( \bigcap C_{\alpha} \right) = \bigcup \left( X \setminus C_{\alpha} \right) = \bigcup O_{\alpha} $$

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$( \implies )$

If $\mathscr{C} := \left\{ C_{\alpha} \ | \ \alpha \in \forall \right\}$ is a set whose elements are closed sets of $X$ and it has the f.i.p., then $\mathscr{O} := \left\{ O_{\alpha} = X \setminus C_{\alpha} \ | \ \alpha \in \forall \right\}$ becomes a set whose elements are open sets of $X$.

Assuming $\displaystyle \bigcap_{\alpha \in \forall} C_{\alpha} = \emptyset$, $$ \bigcup_{\alpha \in \forall} \left( X \setminus C_{\alpha} \right) = X \setminus \left( \bigcap_{\alpha \in \forall} C_{\alpha} \right) = X \setminus \emptyset = X $$ holds, and therefore $X \subset \mathscr{O}$, that is, $\mathscr{O}$ becomes an open cover of $X$. Since $X$ is compact, there exists a finite open cover $\mathscr{O} ' = \left\{ X \setminus C_{i} \ | \ i = 1, 2 , \cdots , n \right\}$ that satisfies $\displaystyle X = \bigcup_{i=1}^{n} \left( X \setminus C_{i} \right)$. Meanwhile, $$ X = \bigcup_{i = 1}^{n} \left( X \setminus C_{i} \right) = X \setminus \bigcap_{i=1}^{n} C_{i} $$ so $\displaystyle \bigcap_{i=1}^{n} C_{i} = \emptyset$. This contradicts the premise that $\mathscr{C}$ has the f.i.p., so it must be that $\displaystyle \bigcap_{\alpha \in \forall} C_{\alpha} \ne \emptyset$.

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$( \impliedby )$

If $\mathscr{O} := \left\{ O_{\alpha} \ | \ \alpha \in \forall \right\}$ is an open cover of $X$, then $\mathscr{C} := \left\{ X \setminus O_{\alpha} \ | \ \alpha \in \forall \right\}$ becomes a set whose elements are closed sets of $X$.

Meanwhile, $$ \bigcap_{\alpha \in \forall} C_{\alpha} = \bigcap_{\alpha \in \forall} \left( X \setminus O_{\alpha} \right) = X \setminus \bigcup_{\alpha \in \forall} O_{\alpha} = X \setminus X = \emptyset $$ so $\mathscr{C}$ does not have the f.i.p., and there exists $\mathscr{C} ' = \left\{ C_{i} \ | \ i = 1, 2 , \cdots , n \right\}$ satisfying $\displaystyle \bigcap_{i=1}^{n} C_{i} = \emptyset$. And $$ X \setminus \bigcup_{i = 1}^{n} O_{i} = X \setminus \bigcup_{i=1}^{n} \left( X \setminus C_{i} \right) = X \setminus \left( X \setminus \bigcap_{i=1}^{n} C_{i} \right) = \bigcap_{i=1}^{n} C_{i} = \emptyset $$ so $\displaystyle X \subset \bigcup_{i=1}^{n} O_{i}$. In other words, there exists a finite subcover $\left\{ O_{1}, O_{2}, \cdots , O_{n} \right\}$, so $X$ becomes compact.

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