Pseudocompactness is continuous image-closed

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)
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
Get more facts about pseudocompact space |Get facts that use property satisfaction of pseudocompact space | Get facts that use property satisfaction of pseudocompact space|Get more facts about continuous image-closed property of topological spaces

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 $g:X\to \mathbb{R}$, $g(X)$ is bounded. A continuous map $f:X\to Y$.

To prove: For any continuous map $h:f(X)\to \mathbb{R}$, $h(f(X))$ 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 $h:f(X)\to \mathbb{R}$. Define $g=h\circ f$. Then, $g:X\to \mathbb{R}$ 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.