Condition for Lipschitz functions and set inclusion

Question

Let $(X,d)$ be a pseudometric space, and for each $A\subseteq X$ and $\varepsilon \geq 0$ define $$ A^\varepsilon := \{x\in X:\exists a\in A \text{ s.t. }d(x,a)\leq \varepsilon\}. $$ Is this set always closed? Denote the closure of $A$ by $\overline A$. Clearly, if $A\subseteq \overline{B^\varepsilon}$ then $$ \sup_{x\in A}f(x) \leq \sup_{x\in B}f(x) +\varepsilon \tag{1} $$ for any 1-Lipschitz function $f:X\to \Bbb R$, that is satisfying $|f(x) - f(y)|\leq d(x,y)$ for all $x,y\in X$. I wonder whether $(1)$ also implies $A\subseteq \overline{B^\varepsilon}$. Any hints are appreciated; I'm also interested in case when $d$ is a metric rather than just a pseudometric.

Answer 1

$A^\varepsilon$ need not be closed. If we take $A = B_r(0)$ in a normed vector space (for example $\mathbb{R}^n$), then $A^\varepsilon = B_{r+\varepsilon}(0)$ is open, but not closed. Even if $A$ is closed, $A^\varepsilon$ need not be closed, take

$$A = \left\{ \left(\varepsilon + 2^{-n}\right)\cdot e_n : n \in \mathbb{N}\right\} \subset \ell^p(\mathbb{N})$$

for your favourite $p\in [1,\infty]$. Then $0 \in \overline{A^\varepsilon} \setminus A^\varepsilon$.

Regarding the containment, if we take

$$f(x) = \operatorname{dist}(x,B) = \inf \left\{ d(x,y) : y \in B\right\},$$

then $f$ is 1-Lipschitz, and

$$\sup_{x\in A} f(x) \leqslant \varepsilon = \sup_{x\in B} f(x) + \varepsilon \implies A \subset \bigcap_{\delta > \varepsilon} B^\delta.$$

For some pseudometrics, the intersection is just $\overline{B^\varepsilon}$, so in these cases the containment follows.

Generally, the containment need not hold. Consider (for a given $\varepsilon > 0$) the (metric) space $(-\infty,0] \cup (\varepsilon,+\infty) \subset \mathbb{R}$ with the (restriction of the) standard metric. Let $B = (\varepsilon,+\infty)$, and $A = B \cup \{0\}$. Then $\overline{B^\varepsilon} = B^\varepsilon > B$, but for every 1-Lipschitz function $f$, we have

$$f(0) \leqslant (\varepsilon + \delta) + f(\varepsilon+\delta) \leqslant \sup_{x\in B} f(x) + \varepsilon + \delta$$

for every $\delta > 0$, so $f(0) \leqslant \sup\limits_{x\in B} f(x) + \varepsilon$, and therefore

$$\sup_{x\in A} f(x) \leqslant \sup_{x\in B} f(x) + \varepsilon.$$