Intersection of Nested Closed Subsets of Compact Space is Non-Empty

Theorem

Let $\struct {T, \tau}$ be a compact topological space.

Let $\sequence {V_n}$ be a sequence of non-empty closed subsets of $T$ with:

$V_{i + 1} \subseteq V_i$

for each $i$.

Then:

$\ds \bigcap_{n \mathop = 1}^\infty V_n \ne \O$

Proof

From Closed Subspace of Compact Space is Compact:

$V_n$ is compact for each $n$.

Aiming for a contradiction, suppose that:

$\ds \bigcap_{n \mathop = 1}^\infty V_n = \O$

Then:

\(\ds V_1\)

\(=\)

\(\ds V_1 \setminus \paren {\bigcap_{n \mathop = 1}^\infty V_n}\)

\(\ds \)

\(=\)

\(\ds \bigcup_{n \mathop = 1}^\infty \paren {V_1 \setminus V_n}\)

De Morgan's Laws: Difference with Intersection

Since each $V_n$ is closed in $T$, from Closed Set in Topological Subspace: Corollary we have:

$V_n$ is closed in $V_1$.

So:

$V_1 \setminus V_n$ is open for each $n$.

So:

$\set {V_1 \setminus V_n : n \in \N}$

is a open cover of $V_1$.

Since $V_1$ is compact, there exists a finite subcover:

$\set {V_1 \setminus V_{n_1}, V_1 \setminus V_{n_2}, \cdots, V_1 \setminus V_{n_j} }$

with:

$n_1 < n_2 < \cdots < n_j$

so that:

$\ds \bigcup_{i \mathop = 1}^j \paren {V_1 \setminus V_{n_i} } = V_1$

We then have, by De Morgan's Laws: Difference with Intersection:

$\ds V_1 \setminus \paren {\bigcap_{i \mathop = 1}^j V_{n_i} } = V_1$

By construction we have:

$\ds \bigcap_{i \mathop = 1}^j V_{n_i} \subseteq V_{n_1} \subseteq V_1$

so:

$\ds \bigcap_{i \mathop = 1}^j V_{n_i} = \O$

Since each $V_{n_i}$ is non-empty, for every $x \in V_{n_j}$, there exists some $1 \le k < j$ such that:

$x \notin V_{n_k}$

But this is impossible since $V_{n_j} \subseteq V_{n_k}$, and so we have reached a contradiction.

So, we must have:

$\ds \bigcap_{n \mathop = 1}^\infty V_n \ne \O$

$\blacksquare$