TITLE: Infimum is a continuous function, compact set
QUESTION [14 upvotes]: Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous map. Show that if $Y$ is compact then the function $g: X \rightarrow \mathbb{R}$ defined by $g(x) = \inf \{f(x,y): y \in Y\}$ is also continuous.
No clue here. Can you please help?

REPLY [2 votes]: Old question, I know, but there’s a simple synthetic approach:
If $f: X × Y → ℝ$ is continuous, then so is the curried version $f^\mathrm{curr}\colon X → C(Y,ℝ)$, where $C(Y,ℝ)$ is the space of continuous functions $Y → ℝ$ with the compact-open topology. Since $Y$ is compact, taking infima is continuous as a map $\inf \colon C(Y,ℝ) → ℝ, f ↦ \inf f$. Thus $\inf ∘ f^\mathrm{curr} \colon X → ℝ$ is continuous as well, which is exactly the function in question.