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:


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

Proof
From Closed Subspace of Compact Space is Compact:


 * $V_n$ is compact for each $n$.

Aiming for a contradiction, suppose:


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

then:

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:


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

From construction we have:


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

so, we have:


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

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


 * $x \not \in V_{n_k}$

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

So, we must have:


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