Proof of Heine-Borel

The Proof Behind the Criterion

In “Heine-Borel Theorem,” the Heine–Borel criterion was stated as a practical test for compactness, and its proof was established earlier in the course. Here we examine a direct open-cover proof of its more demanding direction: a closed bounded subset of \(\mathbb{R}\) is compact. The proof uses the least-upper-bound property to show first that a closed interval has the finite-subcover property, and then uses closedness to pass from an interval to a general set.

Recall that a set is compact when every open cover of it has a finite subcover. Thus, to prove compactness, we must start with an arbitrary open cover and extract finitely many of its members. We cannot assume that the cover is countable or that its sets have any special form. The proof will work for every open cover.

We prove the closed-and-bounded-to-compact direction below. For the converse, we use the earlier results that every compact subset of \(\mathbb{R}\) is bounded and every compact subset of \(\mathbb{R}\) is closed. The aim here is to make the open-cover argument for the first direction explicit.

First, Prove the Interval Case

Let \([a,b]\) be a closed interval with \(a<b\), and let \(\mathcal{U}\) be any open cover of it. Consider the points \(x\in[a,b]\) for which the initial interval \([a,x]\) can be covered by finitely many members of \(\mathcal{U}\). The set of such points is nonempty: a cover member containing \(a\) contains a small interval around \(a\), so it covers an initial segment. The least-upper-bound property then supplies a supremum. If that supremum were short of \(b\), openness of a cover member containing it would allow the finite cover to extend farther, a contradiction.

Proof. If \(a=b\), some member of the cover contains \(a\), and that single member covers \([a,a]\). Now suppose \(a<b\). Define

Choose \(U_0\in\mathcal{U}\) with \(a\in U_0\). Since \(U_0\) is open, there is \(r>0\) such that \((a-r,a+r)\subseteq U_0\). Set \(y=\min\{b,a+r/2\}\). Then \(y>a\), and every point of \([a,y]\) lies in \((a-r,a+r)\). Therefore \([a,y]\subseteq U_0\), so \(y\in S\). In particular, \(S\) is nonempty. It is bounded above by \(b\), so let \(c=\sup S\).

Choose \(U_c\in\mathcal{U}\) with \(c\in U_c\). Openness gives \(r_c>0\) such that \((c-r_c,c+r_c)\subseteq U_c\). We first show that \(c=b\). If \(c<b\), the definition of supremum gives some \(x\in S\) with \(x>c-r_c/2\). Indeed, otherwise \(c-r_c/2\) would be an upper bound for \(S\) smaller than \(c\). Because \(x\leq c\), every point of \([x,\min\{b,c+r_c/2\}]\) lies in \((c-r_c,c+r_c)\), and hence in \(U_c\). A finite collection from \(\mathcal{U}\) covers \([a,x]\), since \(x\in S\). Adding \(U_c\) therefore gives a finite cover of \([a,\min\{b,c+r_c/2\}]\). Since \(c<b\), this right endpoint is strictly greater than \(c\). That contradicts \(c=\sup S\). Thus \(c=b\).

It remains to show that \(b\in S\), since knowing only that \(b\) is the supremum does not by itself show that \([a,b]\) has a finite cover. Choose \(U_b\in\mathcal{U}\) containing \(b\), and choose \(r_b>0\) with \((b-r_b,b+r_b)\subseteq U_b\). By the definition of supremum, there is \(x\in S\) with \(x>b-r_b/2\). The set \([a,x]\) has a finite cover by members of \(\mathcal{U}\), and \([x,b]\subseteq U_b\). Adding \(U_b\) to that finite collection covers \([a,b]\). Thus \(b\in S\), proving the lemma. \(\square\)

From an Interval to a Closed Bounded Set

Let \(K\subseteq\mathbb{R}\) be closed and bounded. If \(K=\varnothing\), it is compact: the empty subfamily is a finite subcover of every cover of the empty set. Suppose \(K\neq\varnothing\). Boundedness gives real numbers \(a<b\) such that \(K\subseteq[a,b]\). The choice can always be made with \(a<b\), even if an initial bound places all of \(K\) at one point.

Now let \(\mathcal{U}\) be an arbitrary open cover of \(K\). Since \(K\) is closed, \(\mathbb{R}\setminus K\) is open. The family consisting of all members of \(\mathcal{U}\), together with \(\mathbb{R}\setminus K\), is an open cover of \([a,b]\): a point of the interval either belongs to \(K\), in which case a member of \(\mathcal{U}\) covers it, or lies outside \(K\), in which case it belongs to \(\mathbb{R}\setminus K\).

By the Closed-Interval Covering Lemma, finitely many sets from this enlarged family cover \([a,b]\). Discard \(\mathbb{R}\setminus K\), if it was selected. The remaining finite collection consists of members of \(\mathcal{U}\), and it still covers \(K\), because \(\mathbb{R}\setminus K\) contains no point of \(K\). We have found a finite subcover of an arbitrary open cover of \(K\); hence \(K\) is compact. This proves the forward direction of Heine–Borel.

Worked Examples: Seeing the Proof in Use

Where Each Hypothesis Enters

The proof separates the roles of boundedness and closedness. Boundedness places \(K\) inside some closed interval \([a,b]\), where the interval lemma applies. Closedness makes \(\mathbb{R}\setminus K\) open, allowing us to fill the part of \([a,b]\) outside \(K\) and then discard it after finding a finite cover. Without boundedness, there may be no enclosing interval. Without closedness, the complement need not be open, so the enlarged family may not be an open cover of the interval.

The supremum step is the core of the interval argument. A finite cover of an initial segment exists near \(a\). If the segment could not be extended all the way to \(b\), a cover member containing its supremum would cover a small interval around that point. One finite cover member can therefore extend the existing finite cover beyond the supposed stopping point. The final endpoint check matters: a supremum need not belong to the set whose supremum it is, so the proof separately establishes that \([a,b]\) itself has a finite cover.

A common gap in this proof is to say that the cover of \(K\) also covers its enclosing interval. It need not. Adding \(\mathbb{R}\setminus K\) is what makes the interval-covering lemma applicable. Another gap is to conclude that a supremum is covered by a finite collection merely because points below it are. The proof closes that gap by using a cover member containing the supremum and explicitly extending the finite cover to the endpoint.