LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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

Statement

Let $n\ge1$ . Every pseudocompact subset $A\subseteq\mathbb{R}^n$ is closed in the Euclidean topology.

Facts & Assumptions

Given: A pseudocompact subset $A\subseteq\mathbb{R}^n$ , with Euclidean metric $d_2$ and norm $\lVert\cdot\rVert_2$ .

[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, $\mathbb{R}^n$ as the set of functions $n \to \mathbb{R}$ , and $d_1$ , $d_2$ , $d_\infty$ are metrics on it).

[L3]

The reverse triangle inequality gives $|\lVert u\rVert_2-\lVert v\rVert_2|\le\lVert u-v\rVert_2$ , and the Euclidean norm is continuous (The finite and reverse triangle inequalities for a norm; and for $n \ge 1$ every norm $N$ on $\mathbb{R}^n$ satisfies $N(x) \le C\lVert x\rVert_1$ 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

1.1

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

assume-contraL1choose

2.1

Define $f:A\to\mathbb{R}$ by $f(x):=1/\lVert x-p\rVert_2$ . This is defined because $p\notin A$ , so every denominator is positive.

step 1.1construct

2.2

For every real $M>0$ , [L1] and [L2] give $x\in A$ with $\lVert x-p\rVert_2<1/M$ ; then $f(x)>M$ . Hence $f[A]$ is unbounded.

step 1.1L1L2

3.1

The function $f$ is continuous on $A$ : at $a\in A$ put $d:=\lVert a-p\rVert_2>0$ . If $\lVert x-a\rVert_2<d/2$ , then [L3] gives $\lVert x-p\rVert_2>d/2$ , and $|f(x)-f(a)| \le \frac{2}{d^2}\lVert x-a\rVert_2.$ Thus a sufficiently small Euclidean ball about $a$ maps into any prescribed real neighbourhood of $f(a)$ .

step 2.1L2L3

4.1

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

step 3.1step 2.2L4discharge-contradiction∎

Depends on

Used by

For a nonempty subset of ℝⁿ with n≥1, compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 125 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Pseudocompact space (standard reference, not scraped)
Heine-Borel theorem (Wikipedia) (standard reference, not scraped)

A pseudocompact subset of Rn\mathbb{R}^nRn is closed