continuity of a function involving an infimum

Question

Let $q(x,y)$ be continuous function on $\mathbb{R}^2$ and $K$ be a compact set of $\mathbb{R}$. Then, define $$ q_K(y) \triangleq \inf_{x \in K} q(x,y). $$ Is $q_K$ a continous function?

Answer 1

First, note that the inf is actually attained, by a standard argument. I'll give a proof in the language of IST.

Fix some standard $y$, and note that $q$ is uniformly continuous on $[y-1, y+1]\times K$. Let $y \approx y'$. We are to show $q_K(y)\approx q_K(y')$.

Suppose $q_K(y)=q(y,x)$ and $q_K(y')=q(y',x')$. I claim that $q(y',x) \approx q(y',x')$. For if not, $q(y',x) \gg q(y',x')$ (i.e. by a non-infinitesimal amount). Then $q(y,x) \approx q(y',x) \gg q(y',x') \approx q(y,x')$ (first $\approx$ by continuity, last $\approx$ by uniform continuity which is needed as $x'$ may not be standard). That's a contradiction to $x$ realising the inf.

Now $q_K(y') = q(y',x') \approx q(y',x) \approx q(y,x) = q_K(y)$ as required.