Pseudocompactness is continuous image-closed

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This article gives the statement, and possibly proof, of a topological space property (i.e., pseudocompact space) satisfying a topological space metaproperty (i.e., continuous image-closed property of topological spaces)
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Statement

Suppose X is a topological space and f:X \to Y is a continuous map. Then, if X is a pseudocompact space, the image f(X) (endowed with the subspace topology from Y) is also a pseudocompact space.

Proof

Given: A topological space X such that, for any continuous map g:X \to \R, g(X) is bounded. A continuous map f:X \to Y.

To prove: For any continuous map h:f(X) \to \R, h(f(X)) is bounded.

Proof: Note first that since f:X \to Y is continuous, so is f:X \to f(X). Henceforth, we think of f as a map from X to f(X).

Consider a continuous map h:f(X) \to \R. Define g = h \circ f. Then, g:X \to \R is a map. Since both h and f are continuous, so is g, and by assumption, we therefore have g(X) bounded. But g(X) = h(f(X)) by definition, so h(f(X)) is bounded, completing the proof.