Let $X$ be a compact space, let $U$ be an open set in $X$, Let $f:U\to [0,1]$ be a continuous map. Prove that the set $$K=\{(x,t): x \in U , 0 \leq t \leq f(x) \} \subset X \times [0,1]$$ is compact.
My attempt
$f(x) \geq t$ implies that $x \in f^{-1}[t,1]$, since $f$ is continuous $f^{-1}[t,1]$ is closed in X, and hence compact. Therefore $f^{-1}[t,1] \times \{t\}$ is compact in $X \times [0,1]$ and $$K=\bigcup_{t\in[0,1]}f^{-1}[t,1] \times \{t\}$$ But arbitrary union of compact sets is not necessarily compact...
Any help is appreciated, Thanks !
Are you sure it holds if $U$ is open? You can take the function $f$ to just take everything to zero. And $X=[0,1]$ and $U=(0,1)$. Then, I don't think $K$ is compact.
Here is a try if $U$ is closed.
Since the product of two compact spaces is compact, we have that the product space is compact. Thus, it suffices to prove that $K$ is closed. Let $(x_n,t_n) \rightarrow (x,t)$, with $(x_n,t_n) \in K$. Then, we have that for each $n$ it holds that $0\leqslant t_n \leqslant f(x_n) $ and thus the same will hold for the limit. But the limit of $t_n$ is $t$ and the limit of $f(x_n)$ is $f(x)$ due to the continuity of $f$. Thus, $0\leqslant t \leqslant f(x) $ and $(x,t) \in K$.