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Let $(X, d)$ be a metric space and let $A \subset X$. For $x \in X$ define $$d(x,A) = \inf\{d(x, y) \mid y \in A\}.$$
Pick out the true statements:

a. $x \mapsto d(x,A)$ is a uniformly continuous function.

b. If $\operatorname{del} A = \{x \in X \mid d(x,A) = 0\} ∩ \{x\in X \mid d(x,X\setminus A) = 0\}$, then $\operatorname{del} A$ is closed for any $A \subset X$.

c. Let $A$ and $B$ be subsets of $X$ and define $d(A,B) = \inf\{d(a,B) \mid a \in A\}$. Then $d(A,B) = d(B,A)$.

I know that (a) is true but no idea about the others. Thanks for your help.

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in c) you just use the fact that you can swap the order of $\inf$ – Ilya Jan 9 '13 at 11:41
del (A) = closure(A) \ closure (X-A) – jim Jan 9 '13 at 11:56
in b> is the question $\operatorname{del} A = \{x \in X \mid d(x,A) = 0\} \setminus \{x\in X \mid d(x,X\setminus A) = 0\}$ or is it $\operatorname{del} A = \{x \in X \mid d(x,A) = 0\} \cap\{x\in X \mid d(x,X\setminus A) = 0\}$ – jim Jan 9 '13 at 13:31
sorry for my mistake. you are right. I have corrcted my fault – user55674 Jan 9 '13 at 13:39
For a see this answer. – Martin Sleziak Jan 9 '13 at 15:27

b> $\operatorname{del}(A) = \overline{{A}} \cap \overline{{X-A}}$ = boundary(A) and boundary(A) is always closed

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