Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm really stuck on this Real Analysis problem, if anyone would mind helping me.

Let $(X,d)$ be a metric space, and let $A$ be a non-empty subset of $X$.

How do you show whether any point $x \in X$ is in the interior, interior of the complement, or boundary of $A$, if the only information you have is $d_A(x) = \inf d(x,a)$ and $d_{A^c}(x)$.

I have been trying to make some arguments using strictly distances but this hasn't been working out. How would you do this?


share|improve this question
add comment

3 Answers

up vote 1 down vote accepted

$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}\newcommand{\bdry}{\operatorname{bdry}}$I gather from your comment to Berci that you have the basic idea but are having a hard time expressing it. It might help to start by proving the observation that for any non-empty $S\subseteq X$ and any $x\in X$, $d_S(x)=0$ if and only if $x\in\cl S$. Then you immediately get that

$$d_A(x)=0=d_{X\setminus A}(x)\quad\text{iff}\quad x\in\cl A\cap\cl(X\setminus A)=\bdry A\;,$$

and you should find it even easier to write down and verify the conditions on $d_A(x)$ and $d_{X\setminus A}(x)$ corresponding to $x\in\int A$ and $x\in\int(X\setminus A)$.

share|improve this answer
add comment

Hint: If $d_A(x)>0$, then $x$ is in the exterior of $A$ (interior of $A^c$). Why?

And, what if $d_A(x)=d_{A^c}(x)=0$?

share|improve this answer
Yes, those are some of the arguments I have been trying to use. If it's greater than zero, the point x is not in the interior because there exists some positive length between a point, a, inside and x outside. If they are both equal to zero it is a boundary point because by definition a boundary point contains at least one element from A and A^c, thus distance zero. I'm still not too sure how to construct this properly though? –  user42538 Oct 11 '12 at 0:21
add comment

Hint: For any $S\subset X$

$$ x\in\mathrm{cl}(S)\Longleftrightarrow d_S(x)=0 $$ $$ X=\mathrm{int}(S)\cup\mathrm{int}(S^c)\cup\partial S $$

share|improve this answer
add comment

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.