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One of the answers given here wrote

  1. A subset $Y \subseteq X$ is called to be somewhere dense if there exists a non-empty open set $U\subseteq X$ such that we have $\overline{Y\cap U}=\overline{U}.$ As one can see, here by some where we actually mean an open set; This attitude seems quite natural since the open sets constitute actually the most fundamental part of a topological space.

  2. A subset $Y \subseteq X$ is called nowhere dense, if it is not the case that it is somewhere dense. It is easy to see that $Y$ is nowhere dense if and only if $\overline{Y}$ does not contain a non-empty open set; the latter is equivalent to the standard definition of a nowhere dense set.

How does one go from (2) to (3)? Shouldn't the case of "not somewhere dense" be

For every open set $U\subset X$ we don't have $\overline{Y \cap U} = \overline{U}?$ Or since $\overline{Y \cap U} \subset \overline{Y} \cap \overline{U} \subset \overline{U}$, this is equivalent to never having $\overline{U} \subset \overline{Y\cap U}?$

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Lemon
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  • Another equivalent formulation: for every non-empty open set $U$ there is a non-empty open subset $V$ of $U$ that misses $Y$. So $Y$ could be called "very avoidable" too. – Henno Brandsma Apr 30 '20 at 22:52

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If $Y$ is nowhere dense in the sense derived from 2., so not anywhere dense, consider $\overline{Y}$: if this contained a non-empty open set $U \subseteq \overline{Y}$, then $$\overline{U \cap Y} = \overline{U \cap \overline{Y}}\tag{1}$$ (as $U$ is open, this holds for any subset $Y$), and the latter equals $\overline{U}$, and so $Y$ would be dense in $U$, contradicting its definition.

If however we know that $Y$ is somewhere dense, say in $U$, we have

$$U \subseteq \overline{U}= \overline{U \cap Y} \subseteq \overline{Y}$$

and $\overline{Y}$ has non-empty interior. So the definitions are equivalent.

Henno Brandsma
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  • So first off, I guess this does boil down to the statement $$\overline{U} \subset \overline{Y \cap U} \iff U \subset \overline{Y}$$ – Lemon Apr 30 '20 at 22:55
  • Next, could you elaborate how $U$ being open implies $$\overline{U \cap Y} = \overline{U \cap \overline{Y}}$$ – Lemon Apr 30 '20 at 22:56
  • I wrote something different. That is first $U \subset \overline{Y} \implies \overline{U} \subset \overline{Y}$. So $$x \in \overline{U} \iff B(x) \cap U \neq \emptyset$$ and $B(x) \cap Y \neq \emptyset$ so $B(x) \cap Y \cap U \neq \emptyset$ is the required condition? – Lemon Apr 30 '20 at 23:00
  • @Hawk That's a standard fact: the left to right inclusion is trivial, and if $x$ is in the right hand side, let $V$ be any open neighbourhood of it: it intersects $U \cap Y$ because $V \cap U$ is also non-empty and must intersect $Y$. etc. – Henno Brandsma Apr 30 '20 at 23:01
  • I just looked at your edited comment. That's more or less what I wrote above yours right? – Lemon Apr 30 '20 at 23:08
  • From right to left that is, because as you said one inclusion is trivial – Lemon Apr 30 '20 at 23:08
  • @Hawk almost, yes. Mine is clearer though, IMO. – Henno Brandsma Apr 30 '20 at 23:09
  • Let me flesh this out. Because $x \in U \subset \overline{Y}$ the condition of being in both closures $\overline{U}$ and $\overline{Y}$ means any neighborhood $B(x)$ must intersect both $U$ and $Y$, but that's exactly means being in $\overline{Y \cap U}$ – Lemon Apr 30 '20 at 23:15
  • Also just a small correction, you mean to write "nowhere dense derived from (3)" right? (2) is somewhere dense – Lemon Apr 30 '20 at 23:22
  • @Hawk it’s the negation of somewhere dense, so derived from 2. And 3 also mentions the alternative so is ambiguous. In your sketch of proof you don’t make it clear where you use openness of $U$, which is essential. – Henno Brandsma May 01 '20 at 06:27