Show that $\bar{A}=\{x \in M | d(x,a)=0\}$

Question

Let $(M,d)$ be a metric space. Let $A$ be an arbitary subset of $M$ and let $x$ be an arbitary point. Define $d(x,A)=\inf \{d(x,y)\mid y \in A\}$. Show that $\bar{A}= \{x \in M \mid d(x,A)=0\}$

How would I approach this? How should I begin?

Answer 1

Show two inclusions, based on your definition of closure as the union of interior and boundary:

If $x \in \operatorname{Int}(A)$ then in particular $x \in A$ and so $d(x,A)$ is the infimum of a set that contains $d(x,x) = 0$ in particular, so $d(x,A) = 0$.

If $x \in \partial A$, pick $r>0$ arbitrarily. Then $B(x,r)$ intersects $A$ by the definition of the boundary, and so there is some point $a_r \in A$ with $d(x,a_r) < r$. So also $d(x,A) \le d(x,a_r) < r$ as well. As $r > 0$ was arbitrary we conclude that $d(x,A) = 0$ as it's below every strictly positive number.

So $\overline{A} \subseteq \{x \in X: d(x,A) = 0\}$.

So suppose $d(x,A) = 0$ for some $x$, and we can assume that $x \notin \operatorname{Int}(A)$ (or we would be done already). This means that for every $r>0$ we have that $B(x,r)$ intersects $X \setminus A$. Also, if $r>0$, some $a_r \in A$ exists such that $d(x,a_r) < r$, because otherwise $\forall a \in A: d(x,a) \ge r$ which implies that $\inf \{d(x,a): a \in A \} \ge r > 0$ as well, a contradiction. So $B(x,r)$ intersects $A$ as well, and as this holds for every $r>0$, $x \in \partial A \subseteq \overline{A}$, as required.

This shows the reverse inclusion.

Answer 2

Let $x\in \bar A$ s.t. $d(x,A)=m>0$. By definition of $\bar A$, there is a sequence $(x_n)\subset \bar A$ s.t. $x_n\to x$ when $n\to \infty $. Let $\varepsilon<m$. Then, there is a $N$ s.t. $d(x_N,x)<m $, which is a contradiction with the fact that $d(A,x)=m$. Therefore, $d(x,A)=0$.