LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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A pseudocompact subset of $R^{n}$ is closed

Statement

Let $n \geq 1$ . Every pseudocompact subset $A \subseteq R^{n}$ is closed in the Euclidean topology.

Facts & Assumptions

Given: A pseudocompact subset $A \subseteq R^{n}$ , with Euclidean metric $d_{2}$ and norm $∥ \cdot ∥_{2}$ .

[L1]

A set is closed if and only if it equals its closure; and $p \in \overline{A}$ means every open neighbourhood of $p$ meets $A$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set).

[L2]

Euclidean open sets are the sets that contain a Euclidean ball about each of their points (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, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

[L3]

The reverse triangle inequality gives $∣ ∥ u ∥_{2} - ∥ v ∥_{2} ∣ \leq ∥ u - v ∥_{2}$ , and the Euclidean norm is continuous (The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ ).

[L4]

Pseudocompactness requires every continuous real-valued function on $A$ to have bounded image (Pseudocompact space: every continuous real-valued function has bounded image).

Proof

technique · contradiction

1.1assume-contraL1choose

Suppose, for contradiction, that $A$ is not closed. By [L1], fix $p \in \overline{A} ∖ A$ .

2.1step 1.1construct

Define $f : A \to R$ by $f (x) : = 1 / ∥ x - p ∥_{2}$ . This is defined because $p \notin A$ , so every denominator is positive.

2.2step 1.1L1L2

For every real $M > 0$ , [L1] and [L2] give $x \in A$ with $∥ x - p ∥_{2} < 1 / M$ ; then $f (x) > M$ . Hence $f [A]$ is unbounded.

3.1step 2.1L2L3

The function $f$ is continuous on $A$ : at $a \in A$ put $d : = ∥ a - p ∥_{2} > 0$ . If $∥ x - a ∥_{2} < d / 2$ , then [L3] gives $∥ x - p ∥_{2} > d / 2$ , and $∣ f (x) - f (a) ∣ \leq \frac{2}{d^{2}} ∥ x - a ∥_{2} .$ Thus a sufficiently small Euclidean ball about $a$ maps into any prescribed real neighbourhood of $f (a)$ .

4.1step 3.1step 2.2L4discharge-contradiction∎

Steps 3.1 and 2.2 contradict pseudocompactness through [L4]. Therefore $A$ is closed.

A pseudocompact subset of $R^{n}$ is closed

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A pseudocompact subset of RnRn is closed

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A pseudocompact subset of $R^{n}$ is closed