Compactness from open covers to subsequences

Real closed sets: the sequence criterion and the open complement

Work on the real line in the usual classical set-theoretic setting, with countable witness selections available. A set $U$ is open when every $y$ in $U$ has some $r>0$ with $(y-r,y+r)$ contained in $U$. We use the Archimedean consequence that $1/n$ tends to zero. For every $K$ subset of the real line, the following conditions are equivalent:

1. Every sequence of points of $K$ that converges in the real line has its limit in $K$.
2. The complement of $K$ is open.

Proof of 2 implies 1. Suppose $x_n$ belongs to $K$ and $x_n$ tends to $x$. If $x$ were outside $K$, openness would give $r>0$ with $(x-r,x+r)$ disjoint from $K$. Convergence gives a threshold after which $|x_n-x|<r$, contradicting $x_n$ in $K$. Therefore $x$ belongs to $K$.

Proof of 1 implies 2. Fix $y$ outside $K$. If no positive-radius interval around $y$ avoided $K$, then for every positive integer $n$ there would be a point $x_n$ in $K$ with $|x_n-y|<1/n$. Choose these witnesses as a sequence. Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$, $|x_n-y|<1/n\le1/N<\varepsilon$, so $x_n$ tends to $y$. Condition 1 would force $y$ into $K$, a contradiction. Hence some interval around $y$ avoids $K$. This holds for every point of the complement, proving it open. The witness-sequence step uses the stated classical setting; this proof is not offered as a theorem about choice-free foundations. If $K$ is empty there is no sequence in $K$, and its complement is the whole open line. If $K$ is the whole line, its empty complement is open vacuously. Both edge cases also satisfy the equivalence.

Learning goals and definitions

A set $K\subseteq\mathbb R$ is compact if every family of open sets whose union contains $K$ has a finite subfamily still covering $K$. A set is closed if it contains all limits of convergent sequences of its points, equivalently if its complement is open. Sequential compactness means every sequence in $K$ has a subsequence converging to a member of $K$.

The empty set is compact, closed and bounded, and sequentially compact vacuously. In proofs using a selected point or an infinite sequence, treat that case separately. We may assume the real completeness property and its nested-interval and Bolzano–Weierstrass consequences from the earlier sequence units.

Theorem: Heine–Borel in the real line

For $K\subseteq\mathbb R$, the following are equivalent: compactness by open covers; closedness and boundedness; sequential compactness.

Compact implies closed and bounded. The intervals $(-n,n)$, $n\ge1$, cover $\mathbb R$, hence $K$. A finite subcover gives a bound. To show closedness, take $y\notin K$. For each $x\in K$, the interval $B(x,|x-y|/2)$ avoids $y$. A finite subfamily covers $K$. Let $\delta$ be the minimum of their positive half-distances to $y$. Every point in the chosen interval centered at $x$ has distance from $y$ strictly greater than $|x-y|/2\ge\delta$. Thus $B(y,\delta)$ misses $K$, showing its complement open. If $K$ is empty this is immediate.

Closed and bounded implies compact. Enclose nonempty $K$ in a closed bounded interval $I_0$. Suppose an open cover of $K$ has no finite subcover. Bisect $I_0$. At least one closed half meets $K$ in a set not finitely coverable; otherwise two finite collections would cover $K$. Choose such a half and repeat. We obtain nested closed intervals $I_n$ with lengths tending to zero, and $K\cap I_n$ not finitely coverable. Each intersection is nonempty; choose $x_n\in K\cap I_n$. The nested-interval property gives a unique common point $p$, and $x_n\to p$ because both lie in $I_n$. Closedness gives $p\in K$. Some member $O$ of the cover contains $p$, and openness gives a ball around $p$ inside $O$. For large $n$, the whole interval $I_n$ lies in that ball, so $O$ alone covers $K\cap I_n$, a contradiction.

Closed and bounded implies sequentially compact. Every sequence in $K$ is bounded, so Bolzano–Weierstrass supplies a convergent subsequence. Closedness retains its limit in $K$.

Sequentially compact implies closed and bounded. If unbounded, choose $x_n\in K$ with $|x_n|>n$; no subsequence can converge to a finite real number, a contradiction. If $p$ is a limit of points in $K$, choose $x_n\in K$ with $|x_n-p|<1/n$. A subsequence converges to a point of $K$ by sequential compactness, but it also converges to $p$; uniqueness of real limits gives $p\in K$. Thus $K$ is closed. This completes all implications.

Corollaries and worked examples

Finite approximation at a prescribed scale. For any $\varepsilon>0$, the balls $B(x,\varepsilon)$ centered at points of compact $K$ form an open cover. A finite subcover gives finitely many sample centers within $\varepsilon$ of every point. Such a finite set is an $\varepsilon$-net. Existence for every scale is called total boundedness; by itself it does not say the set is closed.

Continuous images are compact. If $f:K\to\mathbb R$ is continuous and $(O_i)$ covers $f(K)$ by open sets, the relative inverse images $f^{-1}(O_i)$ cover $K$ and are relatively open. Write each as $K\cap U_i$ with $U_i$ open in $\mathbb R$. Compactness gives finitely many $U_i$ covering $K$, and the corresponding $O_i$ cover the image. Therefore $f(K)$ is compact. If $K$ is nonempty, its compact image is bounded and closed; the supremum and infimum belong to it, recovering the extreme value theorem.

Why boundedness alone fails. The cover $\{(1/n,2):n\ge2\}$ covers $(0,1)$, but every finite subcollection has a positive smallest left endpoint and misses smaller positive points. The set is bounded and even totally bounded, but not compact because zero is a missing limit point.

Graded practice

1. Foundation. Which of $[0,1]$, $(0,1]$, $\mathbb Z$ and $\{0\}\cup\{1/n:n\ge1\}$ are compact in $\mathbb R$?
2. Core. Give an open cover of $\mathbb R$ with no finite subcover.
3. Core. Prove a closed subset of a compact set is compact.
4. Proof. If $K$ is compact and $y\notin K$, prove $\inf_{x\in K}|x-y|>0$ when $K$ is nonempty. Explain why closedness without compactness is enough for positivity but not always for a minimizing sequence proof without extra bounds.

Full solutions

1. The first and fourth are closed and bounded, hence compact. The second is not closed, since $1/n\to0$ outside it. The integers are closed but unbounded, hence not compact.
2. The nested open intervals $(-n,n)$ cover the line, but any finite collection is contained in the largest one and misses all sufficiently large real numbers.
3. If $A\subseteq K$ is closed in $\mathbb R$, it is bounded because $K$ is, and closed by assumption; Heine–Borel proves compactness. For a relatively closed subset, write $A=K\cap C$ with closed $C$ and use the same argument.
4. Continuity of $x\mapsto|x-y|$ gives an attained minimum on nonempty compact $K$. A zero minimum would force $x=y\in K$, impossible, so it is positive. More generally if $K$ is closed and $y\notin K$, openness of its complement directly provides a positive-radius ball avoiding $K$, giving the same lower bound. A minimizing-sequence argument must justify boundedness rather than invoke compactness of an unbounded closed set. Here, eventually $|x_n-y|\le d+1$, where $d=\inf_{x\in K}|x-y|$, so $|x_n|\le |y|+d+1$. A convergent subsequence has its limit in $K$, where the distance attains $d$.

Summary and term definitions

An open cover is a family of open sets containing a set in their union; a finite subcover retains coverage using finitely many members. Sequential compactness keeps subsequential limits inside the set. An $\varepsilon$-net is a finite approximation at one distance scale. The equivalence with closedness and boundedness is special to finite-dimensional Euclidean settings, not a universal definition of compactness in all spaces.