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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

Statement

Let $n \geq 1$ . Every pseudocompact subset $A \subseteq R^{n}$ is bounded for the Euclidean metric.

Facts & Assumptions

Given: A pseudocompact subset $A \subseteq R^{n}$ , where $R^{n}$ has the Euclidean metric $d_{2}$ .

[L1]

The Euclidean norm $x \mapsto ∥ x ∥_{2}$ is continuous from $(R^{n}, d_{2})$ to $R$ (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 $d_{2}$ ).

[L2]

A pseudocompact space has bounded image under every continuous real-valued map (Pseudocompact space: every continuous real-valued function has bounded image).

[L3]

A subset of a metric space is bounded when it is empty or lies in some open ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space); a bounded set of reals has an upper bound (Lower bound, bounded below, bounded set).

Proof

technique · direct

1.1

The restriction $N : A \to R$ , $N (x) = ∥ x ∥_{2}$ , is continuous, because the ambient norm is continuous by [L1] and $A$ has the subspace topology.

1.2

Pseudocompactness gives that $N [A]$ is bounded. If $A \neq \emptyset$ , choose an upper bound $M$ of $N [A]$ ; then $M \geq 0$ because every norm is nonnegative.

L2L3choose

2.1

If $A = \emptyset$ it is bounded. Otherwise every $x \in A$ satisfies $d_{2} (x, 0) = ∥ x ∥_{2} \leq M < M + 1$ , so $A \subseteq B_{d_{2}} (0, M + 1)$ .

step 1.2L3

3.1

Thus $A$ is bounded in both cases.

step 2.1∎

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 · two levels

41 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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

A pseudocompact subset of RnRn is bounded