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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?

Thanks.

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$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}\newcommand{\bdry}{\operatorname{bdry}}$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 $x\in\int(X\setminus A)$.

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

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

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  • $\begingroup$ 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? $\endgroup$ – user42538 Oct 11 '12 at 0:21
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Hint: For any $S\subset X$

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

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