If $(X, d)$ is a metric space, $x$ is in $X$, $S$ is a subset of $X$, we define dist($x$, $S$) := inf { d($x$, $s$) | $s$ is in $S$ }, and define dist($x$, $\emptyset$) := $\infty$.
Question:
Let ($R^3$, $d$) be our metric space with the Euclidean metric.
Is it true that:
"For any subset $S$ of $R^3$ and any $p$ in $R^3 - S$, we have dist($p$, $S$) = dist($p$, bd($S$))"?
Here bd($S$) denotes the boundary of S.
Common definitions of "boundary":https://en.wikipedia.org/wiki/Boundary_(topology)#Common_definitions
Yes, it is true.
Since $\operatorname{bd}S\subset\overline S$, it is clear that $d\left(p,\overline S\right)\leqslant d(p,\operatorname{bd}S)$. Suppose that $d\left(p,\overline S\right)<d(p,\operatorname{bd}S)$. Fix a number $R\in\left(d\left(p,\overline S\right),d(p,\operatorname{bd}S)\right)$ and consider the closed ball $\overline{B_R(p)}$. For each $n\in\mathbb N$, let $s_n\in B_R(p)\cap\overline S$ such that $d(p,s_n)<d\left(p,\overline S\right)+\frac1n$. Since $\overline{B_R(p)}$ is compact, $(s_n)_{n\in\mathbb N}$ has a convergent subsequence and we can assume, without loss of generality, that the sequence $(s_n)_{n\in\mathbb N}$ coverges to some $s$. Then $s\in\overline S$ and $d(p,s)=d\left(p,\overline S\right)$. If $p\notin\operatorname{bd}S$, then $p\in\mathring S$, but then if you pick $r>0$ such that $B_r(s)\subset S$, you can find a point $s^\star$ in the line segment joining $p$ to $s$ such that $s^\star\in S$ and that $d(p,s^\star)<d(p,s)$, which is a contradiction.
Therefore, $d\left(p,\overline S\right)=d(p,\operatorname{bd}S)$ And it should be clear that $d\left(p,\overline S\right)=d(p,S)$.