Let $\{K_i\}_{i=1}^{\infty}$ a decreasing sequence of compact and non-empty sets on $\mathbb{R}^n.$ Then $\cap_{i = 1}^{\infty} K_i \neq \emptyset.$
I heard about a proof that take $x_i \in K_i.$ Then I have constructed a sequence. It is bounded because for every $i$, $x_i \in K_1$ that is bounded. Then there is a subsequence that converges. That is where I stopped. What indeed means that a subsequence converges? And why does the limit of such subsequence is in every $K_i,$ showing that such intersection is not empty.
Thank you!
For each $i\in \mathbb N$ let $x_i\in K_i$, since $K_i$ is non-empty. As the sequence $K_1 \supseteq K_2 \supseteq \dotsb$ is decreasing. It follows $(x_i)_{i\ge 1}$ is contained by $K_1$. By compactness of $K_1$ it follows $(x_i)_{i\ge 1}$ has a converging subsequence $(x_{i_k})_{k\ge 0}$ with limit $x$.
For each $n\in\mathbb N$ the subsequence $(x_{i_k})_{k\ge n}$ starting with $i_n$ is contained by $K_{i_n}$, which is closed. Thus, the limit $x$ is also contained by $K_{i_n} \subseteq \dotsb \subseteq K_1$. As $i_n\to\infty$ for $n\to\infty$, it follows $x\in K_i$ for each $i\in\mathbb N$, that is $x\in\bigcap_{i\in\mathbb N} K_i$.