TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A compact subset of a metric space is closed and bounded

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $K \subseteq X$ be a compact subset (Open cover, subcover, compact metric space, and compact subset of a metric space). Then $K$ is closed in $X$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) and bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

No choice principle is used: both covers below are given by a rule, and the indexed form of A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it returns indices rather than sets.

The converse is false in general. A closed and bounded subset of an arbitrary metric space need not be compact (FALSE: a closed and bounded subset of a metric space is compact); it is exactly in $R^{n}$ that the converse holds (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Facts & Assumptions

Given: A metric space $(X, d)$ and a compact subset $K \subseteq X$ .

[L1]

$K$ is a compact subset exactly when for every set $I$ and every family $(U_{i})_{i \in I}$ of open subsets of $X$ with $K \subseteq ⋃_{i \in I} U_{i}$ there are $n \in N$ and $i_{0}, \dots, i_{n} \in I$ with $K \subseteq U_{i_{0}} \cup \dots \cup U_{i_{n}}$ , or else $K = \emptyset$ (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, Open cover, subcover, compact metric space, and compact subset of a metric space).

[L2]

For $p \neq x$ in $X$ and $r : = d (p, x) / 2$ one has $r > 0$ and $B (p, r) \cap B (x, r) = \emptyset$ (Distinct points of a metric space have disjoint balls around them).

[L3]

Open balls are open, $X$ is open, and a set is closed exactly when its complement is open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

[L4]

A nonempty finite set of reals has a maximum and a minimum, each one of its members (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L5]

A subset is bounded when it is empty or contained in some ball $B (x_{0}, r)$ with $r > 0$ ; and $B (x_{0}, s) \subseteq B (x_{0}, r)$ whenever $0 < s \leq r$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space).

Proof

technique · direct

1.1L3L5

If $K = \emptyset$ then $K$ is bounded by the first clause of the definition, and it is closed because $X ∖ K = X$ is open.

1.2L3L5

Assume from now on that $K \neq \emptyset$ and fix $p \in K$ ; the family $(B (p, r))_{r \in P}$ indexed by the set $P$ of positive reals consists of open sets and covers $K$ , since every $x \in K$ satisfies $d (p, x) < d (p, x) + 1$ and so lies in $B (p, d (p, x) + 1)$ .

2.1L1L4L5step 1.2

The indexed characterisation gives $n \in N$ and positive reals $r_{0}, \dots, r_{n}$ with $K \subseteq B (p, r_{0}) \cup \dots \cup B (p, r_{n})$ ; putting $R : = \max {r_{0}, \dots, r_{n}}$ , a positive real, the balls with common centre $p$ are nested, so $K \subseteq B (p, R)$ and $K$ is bounded.

3.1L2step 2.1

Boundedness being settled, take up closedness: let $q \in X ∖ K$ and for each $x \in K$ put $r_{x} : = d (q, x) / 2$ , which is a positive real because $q \neq x$ , and which satisfies $B (q, r_{x}) \cap B (x, r_{x}) = \emptyset$ .

4.1L1L3step 3.1

The family $(B (x, r_{x}))_{x \in K}$ consists of open subsets of $X$ and covers $K$ , since $x \in B (x, r_{x})$ ; so there are $n \in N$ and $x_{0}, \dots, x_{n} \in K$ with $K \subseteq B (x_{0}, r_{x_{0}}) \cup \dots \cup B (x_{n}, r_{x_{n}})$ .

5.1L4step 4.1

Put $r : = \min {r_{x_{0}}, \dots, r_{x_{n}}}$ , a positive real.

6.1L5step 3.1step 4.1step 5.1

Then $B (q, r) \cap K = \emptyset$ : a point $y$ of the intersection would lie in $B (x_{i}, r_{x_{i}})$ for some $i \leq n$ by step 4.1, and also in $B (q, r) \subseteq B (q, r_{x_{i}})$ by step 5.1, whereas those two balls are disjoint by step 3.1.

7.1L3step 1.1step 2.1step 6.1∎

So every point of $X ∖ K$ has a ball around it inside $X ∖ K$ , that set is open, and $K$ is closed; together with steps 1.1 and 2.1 this proves the theorem.

Remarks

Both conclusions use compactness through the same characterisation. The first cover is by concentric balls of every positive radius, which is what boundedness is about; the second is by balls small enough to keep a fixed outside point away, which is what closedness is about. In each case what compactness returns is a finite list of indices, and a maximum or a minimum of finitely many positive reals then does the rest.

Hausdorffness is what makes the second argument work, and every metric space has it (Distinct points of a metric space have disjoint balls around them). The statement is false for topological spaces without that separation property, which is why the proof cites the separation lemma rather than the metric axioms directly.

A compact subset of a metric space is closed and bounded

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources