Compactness: open covers, sequential, Heine–Borel

In one sentence

In a metric space compactness, sequential compactness, and completeness-plus-total-boundedness are the same thing — and "closed and bounded" is not one of them outside $\mathbb{R}^n$.

Why the exam asks it

Because Heine–Borel is taught as though it were the definition of compactness, and it is a theorem about $\mathbb{R}^n$ specifically. Every year there is an option that applies it in a metric space where it is false. The whole topic is worth learning as: what is true everywhere, what is true in metric spaces, and what is true only in $\mathbb{R}^n$.

The idea

$K$ is compact if every open cover of $K$ has a finite subcover. $K$ is sequentially compact if every sequence in $K$ has a subsequence converging to a point of $K$.

$K$ is totally bounded if for every $\varepsilon > 0$ it can be covered by finitely many balls of radius $\varepsilon$.

The equivalence, and its scope

In a metric space: compact $\iff$ sequentially compact $\iff$ complete and totally bounded.

This is a strong and genuinely useful theorem, and the words "metric space" are part of it. In a general topological space the first two are different properties. For this exam the metric setting is the usual one, so treat the three as interchangeable — but know that the equivalence is a theorem, because options sometimes state it as though it were a definition in a topological space.

What is true in every metric space

• Compact $\Rightarrow$ closed and bounded.
• Compact $\Rightarrow$ complete.
• A closed subset of a compact set is compact.
• The continuous image of a compact set is compact.
• A continuous real function on a compact set is bounded and attains its bounds.
• A continuous function on a compact set is uniformly continuous (Heine–Cantor).
• A continuous bijection from a compact space to a metric space is a homeomorphism — the inverse comes free, which is not true without compactness.

What is true only in $\mathbb{R}^{n}$

Heine–Borel: $K \subseteq \mathbb{R}^n$ is compact $\iff$ $K$ is closed and bounded.

The reverse implication is the one that fails elsewhere, and it fails in the two examples worth memorising.

Any infinite set with the discrete metric. It is bounded (every distance is $0$ or $1$) and closed (everything is), and the cover by singletons has no finite subcover. It is also bounded without being totally bounded, which is exactly the gap Heine–Borel papers over in $\mathbb{R}^n$.

The closed unit ball of $\ell^2$. Let $e_n$ be the sequence with $1$ in position $n$ and $0$ elsewhere. Then for $n \ne m$,

$\|e_n - e_m\| = \sqrt{1^2 + 1^2} = \sqrt2 .$

Every pair is the same fixed distance apart, so no subsequence is Cauchy, so none converges. The ball is closed, it is bounded, and it is not compact. More generally the closed unit ball of a normed space is compact exactly when the space is finite-dimensional — that is Riesz's lemma, and it is worth knowing as a statement even if the proof is not.

The standard non-compactness arguments

To show something is not compact, one of these three always works:

1. Exhibit a cover with no finite subcover. For $(0,1]$, take $U_n = (1/n, 2)$. Any finite subfamily has a largest $n$, hence a smallest left endpoint $1/N > 0$, and misses everything in $(0, 1/N]$.
2. Exhibit a sequence with no convergent subsequence. In $(0,1]$, take $x_n = 1/n$: it converges in $\mathbb{R}$, to a point outside the set, so it has no subsequence converging in $(0,1]$.
3. Show it is not closed, or not bounded, or not complete. Any of the three is enough, since compactness implies all of them.

Where intuition breaks

"Closed and bounded, so compact." The whole topic in one sentence. True in $\mathbb{R}^n$; false in an infinite discrete space and in $\ell^2$. A hypothesis dropped: Heine–Borel's hypothesis is the ambient space.

"Bounded, so totally bounded." An infinite discrete space has diameter $1$ and needs infinitely many balls of radius $1/2$, because each such ball is a single point. In $\mathbb{R}^n$ the two coincide, which is why the distinction feels invented until it is needed.

"Compact, so closed" requires Hausdorff, so it can fail here. It cannot: metric spaces are Hausdorff, so in this subject compact does imply closed. The qualification is real in general topology and irrelevant in a metric space — worth knowing so an option cannot unsettle you with it.

"The continuous preimage of a compact set is compact." Take $f : \mathbb{R} \to \mathbb{R}$ constant $0$. Then $f^{-1}(\{0\}) = \mathbb{R}$. It is the image that inherits compactness, not the preimage; a direction reversed.

"A union of compact sets is compact." Finitely many, yes. $\bigcup_n [0, n] = [0,\infty)$, no.

"An arbitrary intersection of compact sets is compact." True in a metric space: the intersection is closed and sits inside a compact set. A true statement worth recognising, since it appears among false ones.

"$K$ is compact and $f$ continuous, so $f$ attains its supremum" — but $f$ maps into a metric space. Attaining bounds needs a real-valued $f$; into a general metric space there is no order and no supremum to attain. The compact image is still compact.

"Every bounded sequence has a convergent subsequence." Bolzano–Weierstrass is about $\mathbb{R}^n$. In $\ell^2$, the $e_n$ are bounded with no convergent subsequence.

The exam's angle

1. Ask what space you are in. $\mathbb{R}^n$ gives Heine–Borel; anything else does not. This one question resolves most of the topic.
2. If it is not $\mathbb{R}^n$, reach for the two counterexamples: infinite discrete space, and the unit ball of $\ell^2$. Between them they break nearly every false option here.
3. To disprove compactness, produce a sequence. It is usually faster than constructing a cover, and the standard choices ($1/n$ near an excluded endpoint, $e_n$ in a sequence space, $n$ in an unbounded set) cover most cases.
4. Check whether the claim is about the image or the preimage. Compactness and connectedness pass forward through continuous maps; nothing passes backward.
5. Remember what compactness buys you: attained bounds, uniform continuity, a continuous inverse. Options often assert these without the compactness hypothesis.

The night before

• Metric space: compact $\iff$ sequentially compact $\iff$ complete + totally bounded.
• Heine–Borel is $\mathbb{R}^n$ only. Compact $\Rightarrow$ closed and bounded always; the converse is the part that fails.
• Infinite discrete space: closed, bounded, not totally bounded, not compact.
• Closed unit ball of $\ell^2$: $\|e_n - e_m\| = \sqrt2$, no convergent subsequence, not compact. Unit ball compact $\iff$ finite-dimensional.
• Continuous image of compact is compact; the preimage is not.
• Finite unions of compacts are compact; arbitrary unions are not. Arbitrary intersections are.
• Continuous on compact: bounded, attains bounds, uniformly continuous.
• To disprove: a cover with no finite subcover, or a sequence with no convergent subsequence.