I'm stumped on the following question, which is problem 1.3.9 in the book Weak Convergence and Empirical Proceses by van der Vaart and Wellner. It is based on the following notion of asymptotic tightness:
Definition: A sequence of random variables $X_1, X_2, \ldots$ taking values in a metric space with metric $d$ is asymptotically tight if for every $\epsilon > 0$ there exists a compact $K_0$ such that for every $\delta > 0$ we have $\varliminf_{n \to \infty} P(X_n \in K_0^\delta) \ge 1 – \epsilon$ where $K_0^\delta = \{x: d(x, K_0) < \delta\}$ is the $\delta$-enlargement of $K_0$.
For convenience, the definition of uniform tightness is
Definition: A sequence of random variables $X_1, X_2, \ldots$ is uniformly tight if for every $\epsilon > 0$ there exists a compact $K$ such that $P(X_n \in K) \ge 1 – \epsilon$ for every $K$.
The problem is:
Claim: A sequence $X_n$ is uniformly tight if and only if it is asymptotically tight and each $X_n$ is tight.
I cannot seem to prove the non-trivial direction of this claim. The following hint is given:
Hint: Fix $\epsilon > 0$. Take a compact $K_0$ with $\varliminf P(X_n \in K_0^\delta) \ge 1 – \epsilon$ for every $\delta$. Choose $n_1 < n_2 < \cdots $ such that $P(X_n \in K_0^{1/m}) \ge 1 – 2 \epsilon$ for $n \ge n_m$. For $n_m < n \le n_{m+1}$ choose a compact $K_n$ with $P(X_n \in K_0^{1/m} – K_n) < \epsilon$. Now $K = \bigcup_{n = 0} ^ \infty K_n$ is compact and $P(X_n \in K) \ge 1 – 3 \epsilon$.
Now, I understand everything in the hint except for the claim that $K = \bigcup_{n = 0} ^ \infty K_n$ is compact. I assume I'm missing something in how $K_n$ is constructed that makes the countable union compact, since obviously a countable union of compact sets isn't necessarily compact. I also can't seem to find a proof of this claim anywhere other than this book, except a proof in the book Introduction to Empirical Processes and Semiparametric Inference which apparently just copies this hint and leaves the claim that $K$ is compact as "an exercise left to the reader."
EDIT: I think I have a sketch now for compactness, so if anyone could verify it I would be appreciative. The $K_n$ can be chosen so that $K_0 \subseteq K_n \subseteq K_0^{1/m}$ for $n_m < n \le n_{m+1}$. Given an open cover $\{U_\alpha\}$ of $K$, find a finite subcover of $K_0$, $U_1, \ldots, U_K$. Next, find $m$ such that $K_0^{1/m} \subseteq \bigcup_{k = 1} ^K U_k$ (exists because $K_0$ is compact and $\bigcup_{k = 1} ^ K U_k$ is open). Hence $U_1, \ldots, U_K$ is an open cover for $K_n$ for all $n > n_m$. Now, just use compactness of $K_1, \ldots, K_{n_m}$ extend the open cover $K_{n_m + 1}, K_{n_m + 2}, \ldots$ to one which covers all $K_n$.
Best Answer
A simpler way to show the non-trivial direction of the posted claim is the follwoing.
Since the sequence of random variables is asymptotically tight, we have that there is some $n^*$ such that, for all $n>n^*$, $\sup_{k\ge n}P(X_n\notin K)<\epsilon$. Uniform tightness thus follows by combining this result with the fact that the collection $(X_1,...,X_{n^*})$ is finite and the fact that every finite collection of tight random variables is also tight.