I saw this problem in my textbook and I am stuck at a step
Let $C\subset R^d$ and $D\subset R^s$ be closed bounded sets and $f: C \times D \to R $ a continuous function. Show that $x \in C \to \min_D f(x,y)$ is continuous.
My attempt:
(Writing the new function as $g(x)$)
$$ |g(x) - g(z) | = |f(x,\alpha) - f(z,\alpha) + f(z,\alpha) - f(z,\beta)| $$
Where $\alpha$ is the vector in $R^s$ where $f(x,y)$ is maximum and $\beta$ is the vector in $R^s$ where $f(z,y)$ is maximum.
Next I take distance between $x$ and $z$ as $ \lt \delta$ and try to apply triangle inequality but the term $|f(z,\alpha) - f(z,\beta)|$ is problematic. I can't proceed further.
Note that, as $x$ varies minimally, $f$ is uniformly continuous! This is important.
So let $x_0\in C,$ $\varepsilon>0$ and let $\delta>0$ such that, by uniform continuity of $f$ on the closed and bounded set $\bar{B}(x_0,\delta) \times D$, $|f(x,\alpha)-f(y,\beta)|< \varepsilon$ for any $x,y\in B(x_0,\delta)$ and $\alpha,\beta\in D$ such that $||x-y||+||\alpha-\beta||\leq \delta$.
Then, however, for any $y$ with $||x-y||<\delta,$ note that for any $\alpha\in D$ $$ f(x_0,\alpha)+\varepsilon \geq f(y,\alpha)\geq f(x_0,\alpha)-\varepsilon, $$ and, taking minimums on either side of the inequality proves the desired, since $\varepsilon$ was arbitrary.