Prove that $\bigcap_{j\in\mathbb N}B_j\neq\varnothing$

Question

Prove, if $\{B_j\}_{j\in\mathbb N}$ are a collection of closed balls in $\mathbb R^n$ with the condition that for all $i\in\mathbb N$, $B_{i+1}\subseteq B_i$, prove that $$\bigcap_{j\in\mathbb N}B_j\neq\varnothing$$

Answer 1

The closed ball $B_0$ is compact (by Heine-Borel) and if we suppose that $\bigcap_j B_j = \emptyset$ this implies that $\mathbb{R}^n \setminus B_j, j \ge 1$ form an increasing (the $B_j$ are decreasing, so the complements are increasing) open cover of $B_0$ of proper open subsets, so this cover has no finite subcover. This contradicts the compactness so the assumption of an empty intersection was false.

Answer 2

For each $j \in \mathbb{N}$, let $x_j = \inf B_j$. Each $B_j$ is closed and bounded, hence compact, so $x_j \in B_j$. Since for all $j$ we know that $B_{j+1} \subseteq B_j$, we know that $x_{j+1} \geq x_j$, so $(x_j)_{j\in\mathbb{N}}$ is an increasing sequence. Since $(x_j)$ is also bounded, it converges. Let $x = \lim_{j\to\infty}x_j$. Now, fixing $j$, we know that $B_j \supset (x_{j+k})_{k\in\mathbb{N}}$, and therefore $x \in B_j$, because $B_j$ is closed and $x_{j+k} \to x$ when $k\to\infty$. So $x \in B_j$ for each $j$. Thus we can conclude that $x \in \bigcap_{j\in\mathbb{N}} B_j$, implying that $\bigcap_{j\in\mathbb{N}} B_j \neq \varnothing$.