Showing $\{x\in X|f(\{x\}\times[0,1])\subseteq U\}$ is open

Question

Let $f:X\times[0,1]\rightarrow Y$ be a continuous function, and let $U\subseteq Y$ be an open set. Show that the set $V=\{x\in X|f(\{x\}\times[0,1])\subseteq U\}$ is open in $X$.

I don't really know where to start with this problem, and I don't see how it relates to compactness or boundedness, which is what the preceding chapter was about.

Answer 1

Let $x\in V$ and $t\in[0,1]$, let $V_{x,t}$, $U_{x,t}$ such that $x\in V_{x,t}$, $t\in U_{x,t}$ and $f(V_{x,t}\times U_{x,t})\subseteq U$. Now $[0,1]\subseteq U:=U_{x,t_{1}}\cup\cdots\cup U_{x,t_{N}}$, let $W=V_{x,t_{1}}\cap\cdots\cap V_{x,t_{N}}$, now $x\in W\subseteq V$.

Answer 2

The compactness of $[0,1]$ makes the projection $X\times [0,1]\to X$ into a closed map. Now note that the complement of $V$ is the projection of the complement of $f^{-1}(U)$, and therefore closed.