Compactness and Sequential Compactness in Familiar Metric Spaces

Hello! Last lesson used sequences to test whether a function preserves limits. This final lesson in the metric-spaces module uses them to test a property of sets. The central result says that, in a metric space, two apparently different demands—controlling every open cover and controlling every sequence—are equivalent.

You will use that equivalence to recognize compact subsets of familiar spaces and to see why a useful test in Euclidean space does not work everywhere.

---

Two ways to describe compactness

Let $K$ be a subset of a metric space $(X,d)$, equipped with the metric inherited from $X$.

An open cover of $K$ is a family of sets open in $K$ whose union is $K$. The set $K$ is compact if every open cover has a finite subfamily that still covers $K$. In sequential compactness, the starting point is instead a sequence:

$K$ is sequentially compact if every sequence $(x_n)$ in $K$ has a subsequence $(x_{n_k})$ converging to a point in $K$.

The location of the limit matters. For example, $1/n$ is a sequence in $(0,1)$, but its limit $0$ is outside $(0,1)$; no subsequence repairs that problem.

For arbitrary metric spaces, the theorem is

$$\boxed{K\text{ is compact}⟺K\text{ is sequentially compact}.}$$

Some texts use “compact” for the sequential condition and call the open-cover condition the Heine–Borel property. We will distinguish the definitions, then use the theorem to move between them.

---

Why finite subcovers force convergent subsequences

The less obvious part of the forward direction is how an arbitrary sequence could be detected by open sets. The useful objects are its tails: the terms from index $m$ onward. If every tail comes arbitrarily close to the same point $x$, we can choose terms farther and farther along the sequence that converge to $x$.

In Euclidean Space and Metric Spaces, read the first direction of Theorem 8.2.16. Its proof works by contraposition: from a sequence with no subsequence converging in $K$, it builds an open cover with no finite subcover.

Here is a direct version that does not require the sequence terms to be distinct. Take any sequence $(x_n)$ in a compact $K$, and let $F_m$ be the closure in $K$ of its $m$-th tail:

$$F_m={\overline{\{x_n:n\ge m\}}}^K.$$

Each $F_m$ is nonempty and closed in $K$, and the sets are nested:

$$F_1\supseteq F_2\supseteq F_3\supseteq \cdots .$$

They must have a common point. If their intersection were empty, the open sets $K\setminus F_m$ would cover $K$. Compactness would supply finitely many of them that cover $K$. Taking the largest of their indices, say $M$, would mean $K\setminus F_M=K$, contradicting $F_M\ne \varnothing$.

Choose $x$ in every $F_m$. Being in the closure of each tail means that, however far along the sequence we go, some remaining term lies within any prescribed distance of $x$. Inductively choose increasing indices $n_k$ with

$$d(x_{n_k},x)<\frac{1}{k}.$$

Then $x_{n_k}\to x$, and $x\in K$. Thus compactness gives sequential compactness. Notice what the finite-subcover condition accomplished: it prevented all the tails from avoiding a common point.

---

Why convergent subsequences yield finite subcovers

Now suppose $K$ is sequentially compact and $\mathcal{U}$ is an open cover of $K$. Finding a finite subcover requires two ingredients:

1. One radius works throughout the cover. There is an $r>0$ such that, for every $x\in K$, the ball $B_r^K(x)$ lies inside some member of $\mathcal{U}$. The member may depend on $x$. This is the Lebesgue number lemma in the form we need.
2. Only finitely many balls of that radius are needed. For every $r>0$, finitely many radius-$r$ balls centered in $K$ cover $K$. This property is called total boundedness.

Watch the indicated portions of Dr. Bevin Maultsby’s Compact iff Sequentially Compact full proof, Real Analysis II. The first substantial segment proves the common-radius claim; the next uses total boundedness to finish. You can skip the video’s earlier proof of the other direction, which the reading and argument above have covered.

It is worth tracing where sequences enter both ingredients. If $K$ were not totally bounded, some radius $\epsilon >0$ would defeat every finite collection of $\epsilon$-balls. We could then choose $x_1,x_2,\dots$ in $K$ so that each new point lies outside the balls around all earlier points. Distinct terms would remain at least $\epsilon$ apart. No subsequence could be Cauchy, hence none could converge—a contradiction.

For the common radius, suppose no $r>0$ worked. For each $n$, choose $z_n\in K$ whose radius-$1/n$ ball is not contained in any member of the cover. Sequential compactness gives a subsequence $z_{n_k}\to z\in K$. Some $U\in \mathcal{U}$ contains $z$, and because $U$ is open in $K$, a ball $B_{\delta }^K(z)$ lies inside $U$. For sufficiently large $k$,

$$d(z_{n_k},z)<\frac{\delta }{2}\text{and}\frac{1}{n_k}<\frac{\delta }{2}.$$

The triangle inequality puts every point of $B_{1/n_k}^K(z_{n_k})$ inside $B_{\delta }^K(z)$, hence inside $U$. This contradicts our choice of $z_{n_k}$.

Finally, use total boundedness to cover $K$ by finitely many balls of this common radius $r$. Choose one member of $\mathcal{U}$ containing each ball. Those finitely many cover members cover $K$, proving compactness. The common radius connects the given cover to the finite collection of balls; without it, total boundedness alone would not finish the argument.

---

Applying the equivalence

For subsets of ${\mathbb{R}}^m$ with the usual Euclidean metric, the familiar Heine–Borel theorem provides a particularly efficient test:

$$K\subseteq {\mathbb{R}}^m\text{ is compact}⟺K\text{ is closed and bounded}.$$

Read Theorem 8.2.18 in Euclidean Space and Metric Spaces for the sequential reason behind the harder direction. A bounded sequence has bounded coordinate sequences; successively taking subsequences gives convergence in every coordinate. Closedness then keeps the resulting limit in $K$.

The equivalence gives quick diagnoses, but it helps to retain a witness when compactness fails:

• A closed interval $[a,b]$ is compact. Every sequence in it is bounded, so it has a convergent subsequence in $\mathbb{R}$; closedness keeps the limit in $[a,b]$.
• The interval $(0,1)$ is not compact. Its sequence $1/n$, for $n\ge 2$, approaches $0$, and no subsequence can converge to a point of $(0,1)$.
• The set $\mathbb{R}$ is not compact. The sequence $x_n=n$ has no convergent subsequence.

Do not transfer “closed and bounded” to a different metric space without checking. Give $\mathbb{N}$ the discrete metric, where distinct points have distance $1$. The whole space is closed in itself and bounded, yet the sequence $1,2,3,\dots$ has no convergent subsequence: in the discrete metric, a convergent sequence must eventually equal its limit. Equivalently, the cover by all singleton sets has no finite subcover.

The same warning arises even with an ordinary-looking distance. In $C([0,1])$ with the uniform metric from the previous lesson, consider its closed unit ball

$$K=\{f\in C([0,1]):\Vert f{\Vert }_{\infty }\le 1\}.$$

It is closed and bounded, but not compact. The functions $f_n(t)=t^n$ lie in $K$. Every subsequence with increasing indices tends pointwise to $0$ for $t<1$ and to $1$ at $t=1$. That pointwise limit is discontinuous, so no such subsequence can converge uniformly to a member of $C([0,1])$. The Euclidean theorem depends on finite-dimensional structure, not merely on having a metric.

A reliable habit is therefore to state the ambient space and metric before selecting a compactness test. In Euclidean space, closed and bounded is decisive. In a general metric space, use either the open-cover definition or the equivalent demand that every sequence have a subsequence converging inside the set.

---

Takeaways

Compactness turns every open cover into a finite one; sequential compactness turns every sequence into a subsequence converging within the set. These conditions are equivalent for metric spaces. In ${\mathbb{R}}^m$, both reduce to closedness and boundedness, but infinite discrete spaces and spaces of functions show why that shortcut is not universal.

This completes the metric-spaces module. Next, the open sets we have been using will become the starting collection for Borel sigma-algebras, linking metric topology to measurability.