I got stuck on the following question from Bredon Topology and Geometry Chap 1 sec 12.
Suppose $X$ is paracompact. For any open subset $U$ of $X \times [0,\infty)$ which contains $X \times \{0\}$ show that there is a map $f:X\rightarrow (0,\infty)$ such that $(x,y)\in U$ for all $y\leq f(x)$.
The first thoughts on top of my mind is to use a partition of unity, but I have no clue how to proceed. Any hint will be appreciated..
Each $\xi \in X$ admits an open neigbhorhood $U(\xi)$ in $X$ and $t(\xi) > 0$ such that $U(\xi) \times [0,t(\xi)] \subset U$. Now let $\{\phi_\xi\}_{\xi \in X}$ be partition of unity subordinate to the open cover $\{U(\xi)\}_{\xi \in X}$ of $X$. Define
$$f : X \to (0,\infty), f(x) = \sum_{\xi \in X} t(\xi)\phi_\xi(x) .$$ This is well-defined because $\phi_\xi(x) > 0$ only for finitely many $\xi \in X$ and $\phi_\xi(x) > 0$ for at least one $\xi$. It is moreover continuous because each $x \in X$ admits an open neigborhood $V(x)$ such that all but a finite number of the $\phi_\xi$ are $0$. Now let $y \le f(x)$. Let $\xi_1,\ldots,\xi_n$ be the finitely many points such that $\phi_{\xi_i}(x) > 0$. Let $k$ be an index such that $t(\xi_k)$ is maximal among the $t(\xi_i)$. Then $$y \le f(x) = \sum_{i=1}^n t(\xi_i)\phi_{\xi_i}(x) \le \sum_{i=1}^n t(\xi_k)\phi_{\xi_i}(x) = t(\xi_k) \sum_{i=1}^n \phi_{\xi_i}(x) = t(\xi_k) .$$ But $x \in \operatorname{supp} \phi_{\xi_k} \subset U(\xi_k)$, thus $$(x,y) \in U(\xi_k) \times [0,t(\xi_k)] \subset U .$$