Is it true that if $\varepsilon > 0$ and $x \in int(A)$ then $\exists s > 0 \mid d(x,y) \ge \varepsilon + s,\;\forall y \not\in A^\varepsilon$? Let $X=(X,d)$ be a metric space and $A$ be a nonempty subset of $X$ with interior $\overset{\circ}{A}$, closure $\overline{A}$, and boundary $\partial A$. Given $\varepsilon > 0$, define $\varepsilon$-enlargement of $A$ by $A^\varepsilon := \{x \in X \mid d(x,A) \le \varepsilon\}$, where $d(x,A) := \inf_{a \in A} d(x,a)$. Note that $\overline{A} = \{x \in X \mid d(x,A) = 0\}$.
Observe that if $x \in \overline{A}$ and $y \in X\setminus A^\varepsilon$, then $\varepsilon < d(y, A) \le d(x, y) + d(x,A) = d(x,y) + 0$, i.e
$$
d(x,y) > \varepsilon. \tag{1}
$$
One would hope that if $x$ is an interior point of $A$, then the RHS of the bound (1) can be increased.

Question. Is it true that if $x \in \overset{\circ}{A}$ there exists $\delta > 0$ such that $d(x,y) > \varepsilon + \delta$ for every $y \in X\setminus A^\varepsilon$ ?

Affirmative answer for normed inner-product vector space
Suppose $X=(X,d)$ is a normed inner-product vector space. Because $x \in \overset{\circ}{A}$, there exists $\delta > 0$ such that $\{x' \in X \mid d(x',x) < \delta\} \subseteq A$. It is clear that $d(x,\partial A) \ge \delta$. Let $z$ be a point of intersection between $\overline{A}$ and the coord $[x,y]$. Note that we must have $z \in \partial A$ and $d(z,y) > \varepsilon$. By positive collinearity of $x-z$ and $z-y$, we compute
$$d(x,y) = d(y,z) + d(z,x) > \varepsilon + d(z,x) \ge \varepsilon + d(x,\partial A) \ge \varepsilon + \delta.
$$
Thus $d(x,y) > \varepsilon + \delta$.
This motivates the following relaxed question.

Question 2. In case the above question can not be answered affirmatively in general, minimal assumptions can be made about $X$ to alleviate this ?

Edit


*

*It has been observed in the comments that the question doesn't have an affirmative answer in general (example: $X = (-\infty,-1] \cup \{0\} \cup [1,+\infty)$, $A=\{0\}$, $x=0$, $\varepsilon=1$, and $y=-1$). I've noted that all counterexamples seem to be somewhat "pathological". So I'm wondering

*In case $(X,d)$ is a space with the "mid-point property", I've answered the question in the affirmative. See answer below. Such spaces include: normed vector spaces (already proved here), and general complete geodesic spaces.



Question 3. Is there would some general condition (beyond the normed vector-space example) on the space $X$ which would ensure that the question is answered in the affirmative ?

 A: Solution for spaces with the "mid-point property"
Disclaimer: Sorry for answering my own question, but the post below would be too long as a comment or edit to the question.
Let $(X, d)$ have the mid-point property. This means that for every $x,y \in X$ there exists $z \in X$ (called the mid-point of $x$ and $y$) such that $d(x,z) = d(z, y) = d(x,y)/2$. Spaces with the mid-point property include complete geodesic spaces (see Lemma 2.2), etc.

Now, now let $x \in int(A)$. Then there exists $\delta > 0$ such that
$$
\overset{\circ}{B}(x;\delta) \subseteq A, \tag{*}
$$
where $\overset{\circ}{B}(x;\delta) := \{z \in \mathcal X \mid d(z,x) < \delta\}$.

Claim. $d(x,y) > \varepsilon + \delta/2\;\forall y \in X\setminus A^\varepsilon$.

Proof. Let $y \in X\setminus A^\varepsilon$. Set $z_0 := y$, and recursively, for each $k\in \mathbb N$, define $z_{k+1}$ to be the mid-point of $x$ and $z_{k}$. It's easy to see that
$$
d(x,z_k) = 2^{-k}d(x,y) \text{ and }d(x,y) = d(x,z_k) + d(z_k,z_{k-1}) + \ldots + d(z_1,y).
$$
So, by triangle inequality, it holds that
$$
d(x,y) \ge d(x,z_k)+d(z_k,y)\;\forall k \ge 1. \tag{2}
$$
Let $k^* \ge 1$ be the smallest $k$ such that $2^{-k^*}d(x,y) < \delta$ (exists by Archimedian property + well-ordering principle).
Then $d(x,z_{k^*}) = 2^{-k^*}d(x,y) < \delta$, i.e $z_{k^*} \in \overset{\circ}{B}(x;\delta)$. I claim that
$$
d(x,z_{k^*}) \ge \delta/2. \tag{3}
$$
Otherwise, $2^{-(k^*-1)}d(x,y)= 2d(x,z_{k^*}) < \delta$, contradicting the minimality of $k^*$.
Now, by (2) and (3), we have $d(x, y) \ge d(x,z_{k^*}) + d(z_{k^*},y) \ge \delta/2 + d(z_{k^*},y)$. Because $z_{k^*} \in \overset{\circ}B(x;\delta) \subseteq A$ by (1), and $d(y,A) > \varepsilon$ since $y \in X\setminus A^\varepsilon$, it holds that $d(z_{k^*},y) \ge d(A,y) > \varepsilon$. Therefore $d(x,y) > \varepsilon + \delta / 2$. $\quad\quad\quad\Box$
