# Intersection of nested sequence of non-empty compact sets is non-empty (using sequential compactness)

Let $$(X,d)$$ be a metric space, and let $$K_1, K_2, K_3, \ldots$$ be a sequence of non-empty compact sets in this metric space such that $$K_1 \supseteq K_2 \supseteq K_3 \supseteq \cdots$$ Then the intersection $$\bigcap_{n=1}^\infty K_n$$ is non-empty. I am aware of the standard proof of this fact using covering compactness, but I am wondering if there is a proof that uses instead sequential compactness.

My attempt is the following. Since each $$K_n$$ is non-empty, we can pick some point $$x_n \in K_n$$ for each $$n$$ (this requires the axiom of choice). Now consider the sequence $$(x_n)_{n=1}^\infty$$. By the nesting property, we have $$x_n \in K_1$$ for each $$n$$. Since $$K_1$$ is compact, this means there is a convergent subsequence $$(x_{n_j})_{j=1}^\infty$$ which converges to a point $$p \in K_1$$. We will now show that in fact $$p \in K_n$$ for each $$n$$, which would prove that $$p \in \bigcap_{n=1}^\infty K_n$$ (hence, the intersection is non-empty). Let $$n$$ be arbitrary. We have $$n_j \geq j$$ so if $$j \geq n$$ then $$n_j \geq n$$. By the nesting property, this means that $$x_{n_j} \in K_n$$ for all $$j \geq n$$. Thus $$(x_{n_j})_{j=n}^\infty$$ is a sequence of points in $$K_n$$ which converges to $$p$$. Since $$K_n$$ is compact, it is closed, so $$p \in K_n$$.

It seems to me that the proof goes through, but all the proofs I could find online used covering compactness, which made me nervous that somehow using sequential compactness here doesn't work. I would be curious to hear if the proof above goes through (and if not, whether there is another way to use sequential compactness).

My short answer is that I believe your proof is correct. Once you have a sequence of points $$y_i\in K_i$$ which converges (which you have properly shown exists) you just note that this means the point $$y$$ to which it converges must be in $$K_n$$ for any $$n$$ you choose, by exactly the argument you gave. It's quite good.