Prove that $K=\cup_{n=1}^\infty K_n$ is compact

Question

Considering $\mathbb{R}^2$ with the standard metric. Suppose that, for each integer $n\geq 1$, there is a compact set $K_n \subseteq \mathbb{R}^2$ and a point $x_n \in\mathbb{R}^2$ such that:

• $K_n \subseteq B(x_n, 2^{-n})$
• the sequence $(x_n)_{n=1}^\infty$ converges to a point $x \in K$

Prove that $K=\cup_{n=1}^\infty K_n$ is compact.

I know that the infinite union of compact sets is not necessarily compact, so I would like to know where that fails and why the fact that the sequence of the centres of the sets that bound each $K_n$ fixes that. I want to use the Heine-Borel theorem in $\mathbb{R}^n$ and therefore to prove that $K$ is closed and bounded, and have thought to say $K \subset B(x_1, ||x1-x||+1/2)$, but do not see why I can say that the set is closed.

Answer 1

I know that the infinite union of compact sets is not necessarily compact, so I would like to know where that fails and why the fact that the sequence of the centres of the sets that bound each $K_n$ fixes that.

Let's see. So why would an infinite sequence of compact sets not be compact? By the Heine Borel theorem,

• That union could be unbounded, even though each set is bounded. That's very clear, take say $\mathbb R = \cup_{n \geq 1} [-n,n]$. What you see is that $\mathbb R$ is unbounded basically because even though each set is unbounded, there is a point in each set that lies outside any particular fixed bound you give me. For example, you give me $74$ but then $[-75,75]$ is in the union.

The best way of phrasing the above is that $\cup_{i=1}^{\infty} K_i$ won't be compact if there exists a sequence $y_i \in \cup_{i=1}^{\infty} K_i$ such that $\|y_i\| \to \infty$, because that's what you get by contradicting boundedness.

• That union need not be closed, even though each set is closed. That's clear by taking $[0,1) = \cup_{i=1}^{\infty} [0,1-\frac 1{i}]$. The problem here is that if you take a sequence which doesn't converge in the union (or a limit point lying outside the union), then that sequence is something created out of an element in each set.

The best way of phrasing the above is that $\cup_{i=1}^{\infty} K_i$ won't be compact if there exists a sequence $y_i \in \cup_{i=1}^{\infty} K_i$ such that $y_i$ converges to a point outside $K$ (a limit point of $K$ lying outside $K$), because that's what you get by contradicting closedness.

---

So how do you work? You have two tasks :

• Make sure that every sequence is bounded (by the same constant). This will make the whole set bounded.

• Make sure that every sequence is converging to a point inside $K$. This will make the set closed.

But what about the structure of the given set does this?

---

The answer to both questions centers around the fact that $x_i \to x$, where $x_i$ are the centers of the balls.

Because this occurs, the maximal distance of $x_i$ from $x$ is something that is bounded i.e. $\sup_{x \geq 1} \|x_n-x\| = M < \infty$.

For boundedness, one can use the triangle inequality : if $y \in \cup_{i=1}^{\infty} K_i$, and say that it belongs to $K_j$, then $$ \|x-y_j\| \leq \|x-x_j\|+ \|x_j-y_j\| \leq M+2^{-j} \leq M+\frac 12. $$

I believe you attempted this argument, except that you probably assumed that $\sup_{x \geq 1} \|x_n-x\| = \|x_1-x\|$ (because it feels like $x_1$ is the "furthest" from $x$). This need not be true, but the maximum I took is what is to be done, leading to the answer.

---

As for closedness, there's something very nice happening here. See, suppose that you have a "bad" sequence which converges outside $K$. As mentioned earlier, that bad sequence has to have elements from all the $K_i$, so that some kind of "overflow" occurs. What we're doing, by ensuring that $K_i \subset B(x_i,2^{-i})$, is that the possibility of that bad subsequence to go haywire is becoming smaller, because the sets $K_i$ are themselves becoming smaller.

So a bad sequence has to not converge, but it also has to somehow belong to each $K_i$ for some $i$. That makes that bad sequence's possibilities continuously shrink : and because your $K_i$s are going to shrink to a point (which is $x$), it turns out that bad subsequence can only go to $x$!

Let's see how. Suppose that $y_i \in \cup_{i=1}^{\infty}$ is any sequence. Then, it is possible to see that one of the following is true :

• Either, there is some $M>0$ such that $y_i \in \cup_{i=1}^{M} K_i$ for all $i$, OR

• There is a subsequence of $\{y_n\}_{n \geq 1}$, call it $y_{n_k}, k \geq 1$ such that $y_{n_j} \in K_j$ for all $j \geq 1$.

In the first case, obviously $\cup_{i=1}^{M} K_i$ is compact, so $\{y_i\}$ will converge to a point within that set itself, if it converges somewhere.

On the other hand, if we have the subsequence $y_{n_j} \in K_j$ for all $j \geq 1$, then keep in mind that $y_{n_j}$ will converge to the same point as the original sequence $y_n$.

However, because of what I said earlier, we are well within our rights to expect that $x$ is the limit of the $y_{n_j}$, and knowing that, we simply execute the triangle inequality : $$ 0 \leq \|x-y_{n_j}\| \leq \|x-x_{n_j}\| - \|x_{n_j} - y_{n_j}\|\leq \|x-x_{n_j}\|+2^{-n_j}. $$

By the squeeze theorem, $\|x-y_{n_j}\| \to 0$, so $y_{n_j} \to x$. Thus, $y_n$ is a convergent sequence, which has a subsequence that converges to $x$. The only possibility left is that $y_n$ converges to $x$ as well, an easy exercise.

And since we assumed that $x \in K$, as part of the second point, we are able to handle the "bad" sequences by forcing them all to go to the exceptional point $x$.

Note that if $x \in K$ then the bad sequences won't have a limit point and one can get a counterexample. This hopefully explains the question, the principle behind solving it, and how it's actually done, modulo basic some details I leave readers to fill in.

Answer 2

Let $\left\{y_m\right\}$ be a sequence in $K$. If every $K_n$ contains only finitely some $y_m$'s, then choose a $y_{m_1}$ belonging to some $K_{n_1}$. Suppose that we have determined $y_{m_1},y_{m_2},\cdots,y_{m_k}$, as $m_1<m_2<\cdots<m_k$ and $n_1<n_2<\cdots<n_k$. Now, since $\left\{y_m\right\}$ lies in infinitely some $K_n$'s, there must be infinitely some $K_n$'s that do not contain $y_1,y_2,\cdots,y_{m_k}$, to which some of $y_m$'s belong. Therefore, we are able to choose a $y_{m_{k+1}}$ from the $y_m$'s, with $m_{k+1}>m_k$, belonging to a $K_{n_{k+1}}$, with $n_{k+1}>n_k$. Therefore we constructed a subsequence $\left\{y_{m_k}\right\}$of $\left\{y_m\right\}$ converging to $x$.

The other case is quite simple. If a subsequence of $\left\{y_m\right\}$ is contained in a single $K_n$, then it has a convergent subsequence.

Now since every sequence in $K$ has a convergent subsequence, $K$ is compact.