Parametrizing the time an element stays in an open subset

Question

Let $X$ be a topological space (If it helps anything, we can assume $X\subseteq\mathbb{R}^n$ or $X$ being a smooth manifold.) and $U\subseteq [0,1]\times X$ an open subset.

Does there exist a continuous map $\mu\colon X\rightarrow (0,\infty)$, such that if $(0,x)\in U$ holds, then $(t,x)\in U$ holds for all $0\le t\le \mu(x)$?

Answer 1

As already noted in a comment by "1999", this is not generally true as written, even for Euclidean spaces. For example take $X = \mathbb{R}$ and $U = \{ (t, x) \in [0,1] \times X \mid t < |x| \}$.

However, with some minor modifications we get a true statement.

Let $X$ be a perfectly normal space and $U$ an open subset of $[0, 1] \times X$. Then there is a continuous $\mu: X \to [0, 1]$ such that for all $x$ for which $(0, x) \in U$

• $\mu(x) > 0$ and

• $(t, x) \in U$ for all $0 \le t \le \mu(x)$.

This follows almost immediately from the following lemma's.

Lemma 1: Let $X$ be a topological space and $U$ an open subset of $[0, \infty) \times X$. Then $f(x) = \min \{ t \in [0, \infty] \mid (t, x) \notin U \}$ defines a lower semicontinuous $f: X \to [0, \infty]$.

Sketch of proof: $f$ is well-defined because $[0, \infty]$ is compact. For lower semicontinuity, it suffices to prove that if $f(x) > M$, then there is a neighbourhood $V$ of $x$ such that $f(y) > M$ for all $y \in V$.

Note that $f(x) > M$ iff $[0, M] \times \{x\} \subset U$. Since $[0, M] \times \{x\}$ is compact, there are open sets $V \subset X$ and $W \subset [0, \infty)$ such that $[0, M] \times \{x\} \subset W \times V \subset U$. Then for each $y \in V$ we have $[0,M] \times \{y\} \subset U$, hence $f(y) > M$.

Lemma 2: Let $X$ be a perfectly normal topological space and $f: X \to [0, \infty]$ lower semicontinuous. Then there is a continuous $g: X \to [0, 1]$ such that $g(x) = 0$ when $f(x) = 0$ and $0 < g(x) < f(x)$ when $f(x) > 0$.

Sketch of proof: For $n \in \mathbb{N}$ put $F_n = \{ x \in X \mid f(x) \le 2^{-n} \}$. Since $F_n$ is closed there is a continuous $g_n: X \to [0, 2^{-n-1}]$ such that $g_n^{-1}[\{0\}] = F_n$.

Put $g = \sum_{n=1}^\infty g_n$. By uniform convergence, $g$ is continuous, and we have $g^{-1}[\{0\}] = \bigcap_{n=1}^\infty F_n = f^{-1}[\{0\}]$. If $f(x) > 0$ there is a largest integer $n$ such that $f(x) \le 2^{-n}$. Then $x \in F_m$ for all $m \le n$, therefore $$g(x) = \sum_{m=n+1}^\infty g_n(x) \le \sum_{m=n+1}^\infty 2^{-m-1} = 2^{-n-1} < f(x)$$.