What if metric space is replaced by arbitrary topological space, will the result still hold?

Question

Let $(X, d)$ be a metric space with no isolated points, and let $A$ be a relatively discrete subset of $X$. Prove that $A$ is nowhere dense in $X$.

relatively discrete subset of $X$:= A subset $A$ of a topological space $(X,\mathscr T)$ is relatively discrete provided that for each $a\in A$, there exists $U\in \mathscr T$ such that $U \cap A=\{a\}$.

My aim is to prove $int(\overline{A})=\emptyset$. Let if possible $int(\overline{A})\neq \emptyset$. Let $x\in int(\overline{A})$. which implies there exists $B_d(x,\epsilon)\subset \overline A=A\cup A'$.

How do I complete the proof?

What if metric space is replaced by arbitrary topological space, will the result still hold?

Answer 1

Let $x\in\mathring{\overline A}$. Then there is a $r>0$ such that $B_r(x)\subset\overline A$. Since $B_r(x)$ is an open set which is contained in $\overline A$, it contains some element $a\in A$. But then, if $r'=r-d(x,a)$, $B_{r'}(a)\subset B_r(x)$. In particular, $B_{r'}(a)\subset\overline A$. This is impossible, since $A$ is relatively discrete.

I will not answer the question from the title, since you already got an answer in the comments.

Answer 2

With your $x$ and $\epsilon$, $$\overline A\setminus B_d(x,\epsilon/2)$$ is a strictly smaller closed set than $\overline A$, hence cannot contain all of $A$. Pick $a\in A\cap B_d(x,\epsilon/2)$, by which we achieve that $$a\in B_d(a,\epsilon/2)\subseteq A\cap\operatorname{int}(\overline A).$$ By relative discreteness, we find $\delta>0$ such that $B_d(a,\delta)\cap A=\{a\}$. Wlog $\delta\le\epsilon/2$. Now $$ S:=(\overline A\setminus B_d(a,\delta))\cup \{a\}$$ is a closed set with $A\subseteq S\subseteq \overline A$, hence $S=\overline A$. We still have $B_d(a,\delta)\subseteq \overline A=S$, but $B_d(a,\delta)\cap S=\{a\}$, which means that $B_d(a,\delta)=\{a\}$, contrary to the assumption that there are no isolated points.

Answer 3

Suppose $X$ is a crowded $T_1$ space and $D$ is relatively discrete.

Suppose (for a contradiction) that there is some non-empty open set $U \subseteq \overline{D}$

In particular, there is some $d \in D \cap U$ (being in the closure of $D$ means every neighbourhood intersects $D$) and as $D$ is relatively discrete, $\{d\}$ is open in $D$, so there is an open set $U_d$ of $X$ such that $U_d \cap D = \{d\}$.

Now I claim that $U \cap U_d = \{d\}$:

The right to left inclusion is clear, as both open sets contain $d$ and if $x \neq d$ existed in $U \cap U_d$, by $T_1$-ness of $X$ it follows that $U \cap U_d \cap (X\setminus\{d\})$ is an open set containing $x$ that misses $D$ entirely (clearly, as $U_d \cap D = \{d\}$ and $(X\setminus \{d\}) \cap \{d\} = \emptyset$) but $x \in U \subseteq \overline{D}$, so this cannot happen. This shows that indeed $U \cap U_d = \{d\}$, making $\{d\}$ open, but this contradicts in turn that $X$ is crowded (has no isolated points)!

So no such $U$ can exist and $\operatorname{int}(\overline{A}) = \emptyset$.