# Prove that the two definitions of nowhere dense sets are equivalent

I have these two definitions of nowhere dense sets

$X$ is a metric/topological space. Then

$a.$) $A\subset X$ is nowhere dense in $X$ if given any non empty open set $U$ there is a non empty open subset $V\subset U$ such that $A\cap V= \varnothing$

$b.$) $A\subset X$ is nowhere dense iff the interior of closure of $A$ is empty i.e. ${(\bar A)^{\circ}}= \varnothing$

Now I get $a \Rightarrow b$ as follows :

When $a$ is given, if $b$ is not true then there is $x\in X$ s.t $x\in {(\bar A)^{\circ}}$ then there exist $r\gt 0$ s.t $B(x,r)\subset {(\bar A)^{\circ}}$

or, $B(x,r)\subset {\bar A}$

If we take $U=B(x,r)$ here then there is no non empty open subset $V$ of $U$
s.t $A\cap V = \varnothing$. This contradicts $a$ so $b$ must be true.

I cannot figure out $b\Rightarrow a$ Need some help there.

Thanks.

• For $b) \implies a)$, you have a set $A$ whose closure has empty interior. You pick a nonempty open set $U$. What is the largest open subset of $U$ that doesn't intersect $A$? Sep 22, 2015 at 20:25
• Indeed. Now, why is that nonempty? Sep 22, 2015 at 20:30
• ...I don't understand how the first definition could be correct, as it applies to $A=X$, by taking $V=U$.
– Ian
Sep 22, 2015 at 20:31
• If $E \setminus F = \varnothing$, what inclusion do we have? Sep 22, 2015 at 20:35
• Right. And why can that not be? Sep 22, 2015 at 20:37

Suppose a subset $\;A \;$is nowhere dense according to Defn(1). Then $\;\overline{ A}\;$has empty interior. Suppose $\;U\;$is any non-empty open set. Then $\;U\;$cannot be entirely contained in $\;\overline{A}\;.\;$ Therefore $\; U\cap (X-\overline{A})\;$is a non-empty open subset of $\;U\;$disjoint from $\;\overline{A}\;$and hence disjoint from $\;A\;$also. Thus Defn(2) holds true.

Conversely, assume that a subset$\; A \;$is nowhere dense as per Defn(2),then given any non-empty open set $\;V\;,\;$there exists a non-empty open subset $\;W \subseteq V\;$disjoint from $\;A\;.\;\;$As $\;W\;$is open, it is disjoint from $\overline{A}$ also. Thus no non-empty open set can be entirely contained in $\overline{A}.\;\;$ie $\;\overline{ A}\;$has empty interior. Hence Defn(1) holds true. Thus both definitions are equivalent.

$$b)\implies a)$$:

Since $$A$$ is nowhere dense, $$(\overline{A})^c$$ is dense and open. This means that for any open set $$U,\:U\cap (\overline{A})^c\neq\varnothing$$. Let $$x\in U\cap (\overline{A})^c$$. Then $$x$$ has an open neighborhood $$V$$ that $$V\subset U\cap (\overline{A})^c\subset (\overline{A})^c$$, i.e. $$V\cap \overline{A}=\varnothing$$. Thus $$V\cap {A}\subset V\cap \overline{A}=\varnothing$$

• The claim that there is an $x \in V$ and an $r>0$ such that $\emptyset \neq B(x,r) \subset int(A \cap V)$ is wrong. Take $X = \mathbb{R}$, $A = \mathbb{Q}$ and $U = (0,1)$. The intersection $A \cap V$ is nonempty for any open $V \subset U$, but it does not contain any open ball - its interior is actually empty.
– mtg
Apr 22, 2020 at 23:59
• No, $\Bbb{Q}$ is not nowhere dense, but dense because $\overline{\Bbb{Q}}=(\overline{\Bbb{Q}})^o=\Bbb{R}$, while nowhere dense set is $(\overline{A})^o=\varnothing$. Apr 23, 2020 at 0:59
• Yes, every kid knows that. I did NOT write $\mathbb{Q}$ is nowhere dense (this would mean that its closure has empty interior, which obviously is not the case). But $\mathbb{Q}$ itself indeed DOES have empty interior - again: you're wrong and the claim I pointed out is wrong. To repeat: there is NO open ball $B(x,r)$ that would be included in the interior of $A \cap V$ for $A = \mathbb{Q}$ and open $V \subset U := (0,1)$, when $X = \mathbb{R}$.
– mtg
Apr 23, 2020 at 1:00
• You assume $A=\Bbb{Q}$. Also the closure of $\Bbb{Q}$ does not have empty interior. My proof is from a standard text of topology, and so is correct of course. Apr 23, 2020 at 1:03
• Direct us to this standard text, please. Please also read what I write... Nobody denied that the closure of $\mathbb{Q}$ has nonempty interior.
– mtg
Apr 23, 2020 at 1:08