# Compactness, Completeness, Connectedness, Oh my! (Part I)

The Big Three

Well, as the title might suggest, there’s a lot of information about these three topics. I remember them as the Big Three of analysis/topology, as they’re three of the most fundamental (and therefore most used) topics in mathematics. Here’s a summary of the three topics, and why you should (hopefully?) care:

Compactness: Think of compactness as the “closest thing to finiteness” in analysis. It allows us to characterize possibly (usually) uncountable sets in terms of finite (not necessarily disjoint) pieces. That is, we have a sort of bijection between the pieces of a compact set $K$, $\rho: K \to J \subset \mathbb{N}$ where $J$ is finite. For an analyst, this is as close to a finite set as we may come. Note that finite sets are also indeed compact.

Completeness: A complete space is one in which all sequences that want to converge do converge. Intuitively, this means that if “points” (I use this term loosely, because points could represent continuous functions, or categories, etc) in a sequence eventually become arbitrarily close with respect to some sort of metric, then there will, in fact, be a limit point contained in the space to which the sequence converges.

Connectedness: This is probably the most intuitive of the three, as it is simply a characterization of those sets which are, well, connected together. That is, a connected set is like one chunk of silly-putty, and I can’t divide it into two pieces without cutting it in at least one place. Every object that appears to you and I connected in $\mathbb{R}^3$ surely can be proven to be so. It should be noted that connectedness is a relative property. Say I consider $\mathbb{Q} \subset \mathbb{R}$. Then, $\mathbb{Q}$ is disconnected (in fact it is totally disconnected). Now, take $\mathbb{Q}$ as a self-contained metric space. Then, the space $(\mathbb{Q}, d_{\mathbb{Q}})$ is a connected space.

(Pretty Much) Everything You’d Want to Know About Compactness

So this will be a three part post, and the first part will deal exclusively with compactness. I intend to provide most everything one could want to know about compact sets. As mentioned above, compactness is the “closest” thing an analyst has to finiteness. Explicitly, given any open covering, a compact set admits a finite sub-covering. Formally,

Definition 1: Let $K$ be a subset of a metric (or topological) space. We say that $K$ is compact if for every open covering $\mathscr{U} = \{U_i\}_{i \in I}$ and $K \subset \bigcup \limits_{i \in I} U_i$, there exists a finite subset $J \subset I$ so that $K \subset \bigcup \limits_{i \in J} U_i$. That is, a finite sub-covering still covers the set $K$.

So that’s nice. Now what are some examples of compact and non-compact sets? Well, in $\mathbb{R}$ the closed unit interval $[0,1]$ is compact, while the open unit interval $(0,1)$ is not. Why is it not? Consider the open covering $\bigcup \limits_{n \in \mathbb{N}} (\frac{1}{n}, 1-\frac{1}{n})$. Clearly this covers the interval, but if I take only finitely many, it will no longer cover the interval. Ahh, our first hint as to how we may characterize compact sets. Compact subsets of $\mathbb{R}$ cannot be open, because I can always just use a similar argument. Now, what other characteristics do we need for a subset to be compact? Well, is the open ray $(0, \infty)$ compact? (Hint: No, no it is not.) So that means we cannot have a compact set be unbounded. So, it seems that, intuitively, compact sets should be closed and bounded. Luckily, this is true.

Theorem 1: Let $(X,d)$ be a metric space. Let $K$ be a compact subset of $X$. Then, $K$ is closed and bounded.

Proof: We first prove $K$ is bounded. This is simple. Fix $x \in K$. Clearly the balls $\mathscr{B} = \{B_n(x)\}_{n \in \mathbb{N}}$ cover $K$. In fact, they cover $X$. Since $K$ is compact, there must be a finite subset $J = \{1,2, \ldots, N \}$ that still covers $K$. Thus, since $B_1(x) \subset B_2(x) \subset \cdots \subset B_N(x)$, we have that $K \subset B_N(x)$ and therefore $K$ is bounded.

Now, we prove $K$ is closed. There are two fairly simple ways to prove this. I’ll begin with a proof that is similar to that of Munkres’ Analysis on Manifolds. (Drawing a picture on this proof will help, a lot.)

We prove that $X \setminus K$ is open. Take $x \in X \setminus K$. Consider the set of decreasing closed balls $\mathscr{C} = \{C_{\frac{1}{n}}(x)\}_{n \in \mathbb{N}}$. Now, we see that $\bigcap \limits_{n \in \mathbb{N}} C_{\frac{1}{n}} (x) = \{x\}$ and thus (DeMorgan’s Law!) $\Big( \bigcap \limits_{n \in \mathbb{N}} C_{\frac{1}{n}} (x) \Big)^{c} = \bigcup \limits_{n \in \mathbb{N}} C^c_{\frac{1}{n}} (x) = X \setminus \{x\}$ So, since the complement of each closed ball is open, we have an open covering $K \subset \bigcup \limits_{n \in \mathbb{N}} C^c_{\frac{1}{n}} (x)$. Since $K$ is compact, finitely many of these open sets cover $K$, say $\bigcup \limits_{i=1}^{N} C^c_{\frac{1}{n}}(x)$. Now, since $C_1(x) \supset C_\frac{1}{2} (x) \supset \cdots \supset C_\frac{1}{N}(x)$, we have $C^c_1(x) \subset C^c_\frac{1}{2}(x) \subset \cdots \subset C^c_\frac{1}{N} (x)$. This implies $K \subset C^c_\frac{1}{N}(x)$. So, take $N+1$. Clearly then, $C_{\frac{1}{N+1}} (x) \subset C_{\frac{1}{N}}(x)$ and so $C_\frac{1}{N+1}(x) \cap C^c_\frac{1}{N}(x) = \emptyset$. Finally, take $C^{\circ}_\frac{1}{N+1}(x)$ to see that this is an open set containing $x$ completely contained in $X \setminus K$. Since $x$ was arbitrary, $K$ is closed, QED.

Now, the alternate proof relies on the following definition:

Definition 2: A topological space $X$ is Hausdorff if, for any two points $x,y \in X$, there exists disjoint open neighborhoods around each point. That is, there exists $U,V$ so that $x \in U$, $y \in V$ and $U \cap V = \emptyset$.

I claim that any metric space is indeed a Hausdorff Space. This can easily be seen, if I consider two points $x,y$ and take $r= \frac{1}{2}d(x,y)$ and use the balls $B_r(x)$ and $B_r(y)$. So, we have the following proof:

Proof of Theorem 1: Let $(X,d)$ be a metric space. $K \subset X$ compact. We shall show that $X \setminus K$ is open. To see this, fix $x \in X \setminus K$. Now since $X$ is Hausdorff, for each $y \in K$ we have pairings of neighborhoods $y \in V_y$ and $x \in U_y$ so that $V_y \cap U_y = \emptyset$. Therefore, consider the collection $\mathscr{V} = \{V_y\}_{y \in K}$. Naturally, $K \subset \bigcup \limits_{y \in K} V_y$ and so, by compactness, there exist finitely many $V_y$‘s that still cover $K$. Write this covering as $\{V_y (i)\}_{i=1}^{N}$. I claim that $U = \bigcap \limits_{i=1}^{N} U_y(i)$ is disjoint from this finite sub-covering. Well, to see this just choose $z \in U$. Clearly then $z$ is not in any of the members of the finite sub-covering, by our choices of these sets (verify this). Thus, $x \in U \subset X \setminus K$ and so $K$ is closed, QED.

This begs the question: Are sets compact if and only if they’re closed and bounded? Sorry man, this isn’t true. As nice as this would be, it simply isn’t true. “Aw”, you say. “At least give me an example?”

Here’s one: Consider the metric space $(\mathbb{Q}, d_\mathbb{Q})$. The set $S : = \{ q \in \mathbb{Q} : 2 < q^2 < 3 \}$ is closed and bounded in $\mathbb{Q}$.(Yay, another exercise) yet the covering $\{(2+\frac{1}{n}, 3-\frac{1}{n}) \}_{n \in \mathbb{N}}$ has no finite subset that still covers $S$.

So much for that. Well… not quite. It turns out that in the metric space $\mathbb{R}$ a set is compact if and only if it is closed and bounded. But, before I prove this, let’s talk about another characterization of compact sets.

Definition 3: A set $A$ is sequentially compact if, for each sequence $\{x_n\}_{n \in \mathbb{N}} \subset A$, there exists a subsequence $\{x_n\}_{n \in J}$ with $J \subset \mathbb{N}$, so that $\{x_n \}_{n \in J} \to x \in A$.

Since the word compact is literally in the definition, I’d hope that there’s some relation between the two.

Here’s that relation:

Theorem 2: Let $(X,d)$ be a metric space. Then $K \subset X$ is compact if and only if it is sequentially compact.

Before we prove compactness $\implies$ sequential compactness, we state the following lemma that should be proven as an exercise (or, it’s in Baby Rudin, chapter 2 if you’re feeling particularly lazy):

Lemma 1: An infinite subset of a compact set $K$ has a limit point in $K$.

Using this, we prove the first direction:

Proof of Theorem 2: Let $E$ be the range of the sequence $\{x_n\}_{n \in \mathbb{N}} \subset K$. If $E$ is finite, then the sequence is constant after some $N \in \mathbb{N}$  and so the sequence converges in $K$.

Suppose $E$ is infinite. Then, by lemma 1, $E$ has a limit point, call it $x$. Since $x$ is a limit point, the ball of radius $\epsilon$ contains infinitely many points in the sequence (check that if you want), so we may inductively create a subsequence $\{x_k\}$ so that $d(x,x_k) < \frac{1}{k}$. That is to say, $\{x_k\} \to x \in K$, so that $K$ is sequentially compact. QED.

Cool. How about the other direction? For this we need what’s known as the Lebesgue Covering Lemma (at least this is one form, and it will suffice for our proof) and it will be used to prove sequentially compact $\implies$ compact.

Lemma 2 (Lebesgue Covering Lemma): Let $(X,d)$ be a compact metric space. Let $\mathscr{U} = \{U_i\}_{i \in I}$ be an open covering of $X$. Then, there exists $\delta > 0$ so that for each $x \in X$, $B_\delta (x) \subset U_i$ for some $i \in I$.

At first, if you don’t read this closely, it seems trivial. Obviously for each $x \in X$ there’s a $\delta$ so that $B_\delta (x) \subset U_i$ for some $i$. But, the difference here is subtle. This lemma states that there is a single $\delta$ that works for all $x \in X$. Thus, we can have a ball of a single radius that will work for every element.

Proof of Lemma 2: Since $\mathscr{U}$ covers $X$, for each $x \in X$ there exists an $i \in I$ so that $x \in U_i$. As each $U_i$ is open, there exists $\delta(x)$ so that $B_{\delta(x)} (x) \subset U_i$. Consider the collection $\{B_{\frac{\delta(x)}{2}}(x)\}_{x \in X}$. Then, $X \subset \bigcup \limits_{x \in X} B_{\frac{\delta(x)}{2}} (x)$ is an open covering, so there exists a finite sub-covering by compactness of $X$. Call this covering $\{B_{\frac{\delta(x_i)}{2}} (x_i) \}_{i=1}^{N}$. Thus, we see that for an arbitrary $p \in X$, there exists $j$ so that $p \in B_{\frac{\delta(x_j)}{2}}(x_j) \subset U_j$. QED.

Alright, so now we’re ready to prove that sequentially compact implies compact.

Proof of Theorem 2: Assume $X$ is sequentially compact. We will use a proof by contradiction, supposing that $X$ is not compact, and show that there exists some sequence in $X$ that cannot have a convergent subsequence. The proof will rely heavily on the preceding lemma.

Suppose $X$ is not compact. Then, there exists an open covering, say $\mathscr{U}$ so that no finite sub-covering exists. Consider $U_1 \in \mathscr{U}$. Choose $x_1 \in U_1$. By the lemma, there exists $\delta$ so that $B_\delta(x_1) \subset U_1$. By supposition, $U_1$ does not cover $X$. In other words, $X \setminus U_1$ is non-empty, and also not compact. Therefore, we may choose $U_2 \in \mathscr{U}$ so that $U_2 \subset X \setminus U_1$. Pick $x_2 \in U_2$. Again, by the lemma, $B_\delta (x_2) \subset U_2$. Proceed to construct this sequence inductively. Namely, $X \setminus \bigcup \limits_{i=1}^{k-1} U_i$ is non-empty and not compact,

so we may choose $U_k \subset \Big( X \setminus \bigcup \limits_{i=1}^{k-1} U_i \Big)$ and pick $x_k \in U_k$ and look at $B_\delta (x_K) \subset U_k$. Now, what have we really done here? We’ve created a sequence so that $d(x_i,x_j) > \delta$ for all $i \neq j$. Thus, no matter how large we take $N$, two elements will always be at least a distance $\delta$ away from each other. Such a sequence (as you can check as a short exercise) cannot have a convergent subsequence. That is to say, $X$ is not sequentially compact, a contradiction. We conclude the theorem. QED.

Yay. So now we know that, in any metric space (not in any topological space) the notion of compactness is equivalent to the notion of sequential compactness. Now, we can return to the issue of a set in $\mathbb{R}$ (equivalently $\mathbb{R}^n$) being compact if and only if it is closed and bounded. For this, we’ll use the equivalence of sequential compactness. Here’s (half) of a theorem:

Theorem 3: Let $(X,d)$ be a metric space. Let $K$ be a sequentially compact subset of $X$. Then, $K$ is closed and bounded.

The proof of this is a simple exercise. That is, note that, for any limit point $x$, we may construct a sequence that converges to $x$. But then, since $K$ is sequentially compact, there is a subsequence that converges to $y \in K$. By uniqueness of limits, $x=y \in K$ so that $K$ contains its limit points and is therefore closed. To see that it is bounded, suppose not and note that we could construct a sequence that has no convergent subsequence.

I assume that the reader is familiar with the Bolzano-Weierstrass theorem, stating that all bounded subsequences of $\mathbb{R}^n$ have a convergent subsequence. This gives the following theorem:

Theorem 4: Let $K \subset \mathbb{R}^n$ be closed and bounded. Then, $K$ is sequentially compact.

How to prove this? Well, since $K$ is bounded, it has a convergent subsequence by Bolzano-Weierstrass. Since it is closed, this subsequence converges in $K$. So, now we see that sequentially compact and closed and bounded are equivalent statements in $\mathbb{R}^n$. Let’s sum up what we’ve done so far:

• Compactness implies closed and bounded in all metric spaces.
• Compactness is equivalent to sequential compactness in all metric spaces.
• Sequential compactness implies closed and bounded in all metric spaces.
• Sequential compactness is equivalent to closed and bounded in $\mathbb{R}^n$.

Therefore, in $\mathbb{R}^n$ we see compactness $\Leftrightarrow$ sequential compactness $\Leftrightarrow$ closed and bounded. Thus, compactness $\Leftrightarrow$ closed and bounded in $\mathbb{R}^n$. This equivalence is known as the Heine-Borel theorem, which we have proven above. Ok so that’s pretty right? Now instead of checking every open cover of a subset of $\mathbb{R}^n$, I just need to show that the subset is both closed and bounded and therefore compact.

Before I conclude this post, I’d like to prove just two more things about compactness. These are the fact that closed subsets of compact sets are themselves compact, and the fact that compact sets with the finite intersection property (FIP) have non-empty intersection.

Theorem 5: Let $(X,d)$ be a compact metric space. Let $K \subset X$ be closed. Then, $K$ is also compact.

Proof: We need show that every open covering of $K$ admits a finite sub-covering. Let $\{U_\alpha\}_{\alpha \in A}$ be an open covering of $K$. Then, $\bigcup \limits_{\alpha \in A} U_\alpha \cup (X \setminus K)$ is an open covering of $X$. The rest of the proof is left to the reader, which should be fairly straightforward. QED.

Finally, we have the following theorem:

Theorem 6: Let $(X,d)$ be metric space. Let $\{K_\alpha\}_{\alpha \in A}$ be a family of compact sets so that $\bigcap \limits_{\alpha \in J} K_\alpha \neq \emptyset$ with $J$ is finite. That is, finite intersections are not empty. Then, $\bigcap \limits_{\alpha \in A} K_\alpha \neq \emptyset$.

Proof: Suppose that this intersection is empty. That is, $\bigcap \limits_{\alpha \in A} K_\alpha = \emptyset$. So, $\Big( \bigcap \limits_{\alpha \in A} K_\alpha \Big)^{c} = X = \bigcup \limits_{\alpha \in A} K^c_\alpha$ is an open cover of some $K$ in the collection. Since $K$ is in the collection, it is compact, and thus admits a finite sub-covering, say $K \subset \bigcup \limits_{\alpha \in J} K^c_\alpha$. But then, $K \cap \bigcap \limits_{\alpha \in J} K_\alpha = \emptyset$ contradicting the FIP hypothesis. Thus, this intersection is non-empty. QED.

I think that’s all for now, as that provides a fairly exhaustive summary of compactness and it’s basic properties. In future posts, I’ll discuss things like uniform continuity on compact sets, and other things of this nature. If you take away anything from this post, it’s how to characterize compact sets, as well as understanding that compactness is the next best thing to finiteness.

Cheers,

J.T.