Exchanging lim and inf when rewriting $\inf\{t>0:\forall ε>0,S_t\cap B(x_0,ε)\subset S\cap B(x_0,ε)\}$

Question

How can I prove that $$\inf\{t>0: \forall ε>0, S_t \cap B(x_0,ε) \subset S\cap B(x_0,ε)\} \\= \lim_{ε \to 0}\inf\{t>0: S_t \cap B(x_0,ε) \subset S \cap B(x_0,ε) \},$$ where $S$, $S_t\ (t \ge 0)$ are closed sets in $\mathbb{R}^n$.

Answer 1

The question assumes that the assertion is true, which it is not. Let me give a one-dimensional counterexample. Let $x_0=0$, $S=[0,1] \cup \{2\}$, $S_t = \mathbb{R}$. Then: $$\inf\{t>0: \forall ε>0, [-ε,ε] \subset S \cap [-ε,ε]\} = \infty,$$ since the condition is not true for $ε>1$. However, $$\lim_{ε \to 0}\inf\{t>0: [-ε,ε] \subset S \cap [-ε,ε] \} = 0.$$ If you dislike the convention that $\inf \emptyset = \infty$, you can take $S_1 = S$ (while keeping the other $S_t$ equal to $\mathbb{R}$) to get $1$ instead of $\infty$ for the first expression.