int$(A) \subseteq$int$(A')$ and int$(A) \subseteq A'$

Question

In which metric spaces is it true that int$(A) \subseteq$int$(A')$ ? (I know it is true in $\mathbb R$)

Moreover in which metric spaces is it true that int$(A) \subseteq A'$ ? (I know it is true in $\mathbb R$)

$A'$ is the derived set of $A$ http://en.wikipedia.org/wiki/Derived_set_(mathematics)

Answer 1

Both conditions are equivalent to the metric space not having isolated points. More specifically, the following holds:

Theorem: Let $(M,d)$ be a metric space. Then the following are equivalent:
$(a)$ $M$ does not have isolated points;
$(b)$ For every $A\subseteq M$, $\operatorname{int}A\subseteq A'$;
$(c)$ For every $A\subseteq M$, $\operatorname{int}A\subseteq\operatorname{int}A'$.

Proof. $(a)\Rightarrow(b)$. Suppose $M$ does not have isolated points, and let $A\subseteq M$ and $x\in \operatorname{int}(A)$. Then there exists a neighbourhood $U$ of $x$ (an open ball, if you want) such that $U\subseteq A$. Given any other neighbourhood $V$ of $x$, since $M$ does not have isolated points, then there exists some $y\in V\cap U\subseteq V\cap A$, $y\neq x$. Thus, $x\in A'$.

$(b)\Rightarrow (c)$. Let $A\subseteq M$. Then $\operatorname{int}A\subseteq A'$. Apply $\operatorname{int}$ to both sides and obtain $\operatorname{int}A=\operatorname{int}\operatorname{int}A\subseteq\operatorname{int}A'$.

$(c)\Rightarrow (a)$. Let's prove the contrapositive: Suppose $M$ has an isolated point $x$. Then the set $A=\left\{x\right\}$ has $\operatorname{int}A=A\neq\varnothing$, but $A'=\varnothing$, so $(c)$ does not hold.Q.E.D.

Obviously, the same proof applies to general topological spaces.