Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.