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

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?

• If $X$ has indiscrete topology and $A$ is a singleton set, then $A$ is discrete, but $\overline A=X$ is a nonempty open set – Hagen von Eitzen Oct 21 at 10:41
• Thank you for the counter example. How do i complete the proof? Can you please help me? – Math geek Oct 21 at 10:59
• @Mathgeek the result holds in $T_1$ spaces, I believe. – Henno Brandsma Oct 21 at 11:29

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 $$\operatorname{int}(\overline{A}) = \emptyset$$.

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.

• How $B_{r'}(a)\subset \overline A$ contradicts to relative discreteness? – Math geek Oct 21 at 15:12
• @Mathgeek Since $B_{r'}(a)\subset\overline A$ and since $a$ is not an isolated point of $A$, every ball centered at $a$ will contain an element of $A$ distinct from $a$. But, since $A$ is relatively discrete, there should be some open ball centered at $a$ such that the only element of $A$ that it contains is $a$ itself. – José Carlos Santos Oct 21 at 15:18
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.