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A subset of $R$ is compact if and only if it is closed and bounded

Statement

Let $K \subseteq R$ . Then $K$ is compact (Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset) if and only if $K$ is closed (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen) and bounded (Lower bound, bounded below, bounded set).

This is the Heine-Borel theorem in the form used everywhere below. The forward implication is A compact subset of $R$ is closed and bounded and spends no completeness, only the Archimedean property and the existence of maxima of finite sets; the backward implication rests on Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact and therefore on the completeness of $R$ , and the remarks below record where it fails without completeness.

Facts & Assumptions

Given: A subset $K \subseteq R$ .

[L1]

Open cover, finite subfamily and compactness; the empty subfamily covers $\emptyset$ (Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset).

[L2]

A compact subset of $R$ is closed and bounded (A compact subset of $R$ is closed and bounded).

[L3]

Every closed bounded interval $[ℓ, u]$ with $ℓ \leq u$ is compact (Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact).

[L4]

$K$ is closed exactly when $R ∖ K$ is open (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L5]

$K$ is bounded exactly when there are $ℓ, u \in R$ with $ℓ \leq y \leq u$ for every $y \in K$ (Lower bound, bounded below, bounded set).

[L6]

$[ℓ, u] = {z \in R : ℓ \leq z \leq u}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct

1.1L2

If $K$ is compact then $K$ is closed and bounded, which is [L2]; this is the forward implication.

1.2assume-hypL1

For the backward implication assume $K$ is closed and bounded. If $K = \emptyset$ then every open cover of $K$ admits the empty subfamily as a finite subcover, so $K$ is compact.

1.3assume-hypL5L6choose

Assume moreover $K \neq \emptyset$ ; fix $s \in K$ and, by [L5], reals $ℓ, u$ with $ℓ \leq y \leq u$ for every $y \in K$ . Then $ℓ \leq s \leq u$ , so $ℓ \leq u$ , and $K \subseteq [ℓ, u]$ by [L6].

2.1step 1.3L1L4

Let $U$ be an open cover of $K$ and put $W : = U \cup {R ∖ K}$ . Every member of $W$ is open, since $R ∖ K$ is open by [L4], and $W$ covers $[ℓ, u]$ : a point of $[ℓ, u]$ either lies in $K$ , hence in some member of $U$ , or lies outside $K$ , hence in $R ∖ K$ .

3.1step 2.1L1L3L6

By [L3] the interval $[ℓ, u]$ is compact, so some finite subfamily ${W_{0}, \dots, W_{p}}$ of $W$ covers $[ℓ, u]$ , where the case of an empty subfamily is possible only when $[ℓ, u] = \emptyset$ , which is excluded by $ℓ \leq u$ . Put $V : = {W_{i} : W_{i} \in U}$ , a finite subfamily of $U$ . Then $K \subseteq ⋃ V$ : a point $y \in K \subseteq [ℓ, u]$ lies in some $W_{i}$ , and $W_{i}$ cannot be a member of $W$ outside $U$ , because the only such member is $R ∖ K$ and $y \in K$ ; so $W_{i} \in U$ and $W_{i} \in V$ .

4.1step 1.1step 1.2step 3.1L1∎

Every open cover of a nonempty closed bounded $K$ therefore has a finite subcover, so such a $K$ is compact; together with the empty case of step 1.2 this proves the backward implication, and step 1.1 is the forward one.

Remarks

A closed subset of a compact set is compact. If $C \subseteq K$ with $K$ compact and $C$ closed, then $C$ is bounded, being a subset of a bounded set, and closed by hypothesis, so it is compact by the theorem. The corresponding statement for arbitrary subsets is false: $(0, 1) \subseteq [0, 1]$ is bounded and not compact ( $Z$ is closed and not compact, and $(0, 1)$ is bounded and not compact: neither hypothesis of Heine-Borel can be dropped ↗).
Both hypotheses are needed and neither implies the other. $Z$ is closed and not compact, and $(0, 1)$ is bounded and not compact: neither hypothesis of Heine-Borel can be dropped ↗ exhibits a closed set that is not bounded and a bounded set that is not closed, and neither is compact.
What the theorem is not. It characterises compactness for subsets of $R$ . The two halves are of very different strengths: the forward half is elementary and general, while the backward half rests on the completeness of $R$ and fails over $Q$ (FALSE: in every ordered field a closed bounded set is compact, so Heine-Borel needs no completeness, witnessed by ${q \in Q : q \geq 0, q^{2} < 2}$ is closed and bounded in $Q$ and is not compact ↗). Nothing here licenses "closed and bounded implies compact" in any other setting.

A subset of $R$ is compact if and only if it is closed and bounded

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A subset of RR is compact if and only if it is closed and bounded

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A subset of $R$ is compact if and only if it is closed and bounded