Nested Compact Set Property

If:

K_{1} \supseteq K_{2} \supseteq \dots

are nested, compact, nonempty sets. Then:

⋂_{n = 1}^{\infty} K_{n} \neq \emptyset

Sketch of the proof:

Notice $\forall n \in N$ then let $x_{n} \in K_{n}$ . Note $x_{1} \in K_{1}$ , $x_{2} \in K_{2} \Rightarrow K_{1}$ , $x_{3} \in K_{3} \Rightarrow x_{3} \in K_{2} \Rightarrow x_{3} \in K_{1}$ , and so on. Then each $x_{n} \in K_{1}$ . Therefore, then $\exists$ a convergent subsequence $(x_{n_{i}})$ where $x = lim_{i \to \infty} x_{n_{i}} \in K_{1}$

Now $x_{n_{1}} \in K_{n_{1}} \subseteq K_{1}$ since $n_{1} \geq 1$ . Similarly $x_{n_{2}} \in K_{n_{2}} \subseteq K_{2}$ since $n_{2} \geq 2$ . In general $x_{n_{i}} \in K_{i}$ for all $i$ since $n_{i} \geq i$ . Now:

x = lim_{i \to \infty} x_{n_{i}} = (x_{n_{2}}, x_{n_{3}}, \dots) \in K_{2}

We can repeat this process, where the next step shows:

x = lim_{i \to \infty} x_{n_{i}} \in K_{i}

so $x \in K_{i}$ for all $i$ so then their intersection is non-empty.

Proof

For each $n \in N$ pick a point $x_{n} \in K_{n}$ . Because the compact sets are nested then $(x_{n}) \subseteq K_{1}$ and by the definition of compactness then there is a convergent subsequence $(x_{n_{k}})$ whose limit $x \in K_{1}$ .

But you can repeat this process for each $K_{n}$ . To show this use a particular $n_{0} \in N$ so the terms in the sequence $(x_{n}) \subseteq K_{n_{0}}$ when $n \geq n_{0}$ , ignoring the finite number of terms at the start. Thus then the same subsequence $(x_{n_{k}}) \subseteq K_{n_{0}}$ , so then $x$ is a limit of each $(x_{n_{k}}) \subseteq K_{n_{0}}$ . Since $n_{0}$ was arbitrary then $x \in ⋂_{n = 1}^{\infty} K_{n}$ .

☐

This idea is similar to the Nested Interval Property, except it works with generic sets.