# Understanding of nowhere dense sets

The following is the definition of nowhere dense sets in Wikipedia.

In topology, a subset $A$ of a topological space $X$ is called nowhere dense (in $X$) if there is no neighborhood in $X$ on which $A$ is dense.

It is also said in this lecture note that a nowhere dense set is a set $E$ which is not dense in any ball.

I don't quite understand what "not dense in any ball" means unless I can find several essentially different counterexamples.

Considering ${\mathbb R}$ with the standard topology, I am thinking about examples which is NOT a nowhere dense set. Here are my questions:

What can be an example such that the set $E$ is dense in any ball?

What can be an example such that the set $E$ is dense in some balls while it is not in others?

I'm sure what "$E$ is dense in a ball" means. Does it mean $\overline{E}$ is a ball or $\overline{E}$ contains a ball or $\overline{E}$ is contained in a ball?

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For the final part: it means that the (relative) closure of $E \cap B$ (here $B$ is a ball) inside $B$ is all of $B$. – Dylan Moreland Aug 17 '11 at 23:10
You may want to read about the Cantor Set for an interesting non-trivial example. – t.b. Aug 17 '11 at 23:14
After giving you an example you're not asking about, think about $\mathbb{Q}$ for a set that is dense in any ball of $\mathbb{R}$ and $\mathbb{Q} \cap \bigcup_{n \in \mathbb{Z}} [2n,2n+1]$ for a set that is dense in some balls but not in others. – t.b. Aug 17 '11 at 23:22
@Dylan: +1. It makes sense now. Could you give me a reference for your statement? Since I don't see that in the basic topology textbook I learned. – Jack Aug 17 '11 at 23:27
@Jack: No, you cannot control the number of balls in which the set is dense. If $E$ is dense in some ball $B$, then $E$ is automatically dense in every ball that is a subset of $B$, and there are infinitely many such balls. – Brian M. Scott Aug 18 '11 at 0:02

Assume that $E$ and $B$ are subsets of a topological space $X$. By definition the set $E$ is dense in $B$ iff the set $E\cap B$ is dense in the subspace $B$ iff $\operatorname{cl}_B(E\cap B)=B$. Using the equality $\operatorname{cl}_B(E\cap B)=\overline{E\cap B}\cap B$ and elementary observations we obtain the equivalences $$\operatorname{cl}_B(E\cap B)=B\ \Leftrightarrow\ \overline{E\cap B}\cap B=B\ \Leftrightarrow\ \ B\subseteq\overline{E\cap B}\cap B\ \Leftrightarrow\ B\subseteq\overline{E\cap B}\ \Leftrightarrow\ B\subseteq\bar{E}.$$ Hence $E$ is dense in $B$ iff $B\subseteq\bar{E}$.

By definition $E$ is nowhere dense in $X$ iff there is no nonempty open subset (equivalently neighborhood) of $X$ in which $E$ is dense. Using the observations above, $E$ is nowhere dense in $X$ iff $\bar{E}$ does not contain any nonempty open subset of $X$ iff $\operatorname{int}\bar{E}=\emptyset$.

Indeed, in the metric setting, $E$ is nowhere dense iff there is no ball in which $E$ is dense, since any nonempty open subset of a metric space contains a ball.

In any metric space the condition of a subset being dense in every ball is equivalent to being dense in the space, because every point of a metric space is contained in a ball. Thus any dense subset of $\mathbb{R}$, for example $\mathbb{R}$, $\mathbb{R}\setminus\{0\}$ or $\mathbb{Q}$, is an example to your first question. Regarding your second question, the ball $(-1,1)\subseteq\mathbb{R}$ is dense in itself but not dense in any ball that is not contained in $(-1,1)$.

"$E$ is dense in a ball" does not mean "$\bar{E}$ is a ball". For example $(-1,1)\cup\{2\}$ is dense in the ball $(-1,1)$ but $\bar{E}=[-1,1]\cup\{2\}$ is not a ball. However, "$\bar{E}$ is a ball" implies that "$E$ is dense in a ball": if $\bar{E}$ is a ball, then $E$ is dense in the ball $\bar{E}$.

"$E$ is dense in a ball" means "$\bar{E}$ contains a ball", since according to our observations above $E$ is dense in a set $B$ iff $B\subseteq\bar{E}$.

"$E$ is dense in a ball" does not mean "$\bar{E}$ is contained in a ball". For example $\mathbb{R}$ is dense in the ball $(-1,1)$ but no ball contains its closure. On the other hand $\bar{\emptyset}=\emptyset$ is contained in every ball but $\emptyset$ is dense in no ball.

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I don't quite understand the Answer from "LostInMath". The Answer say $B\subset \overline{E\cap B}\Leftrightarrow B\subseteq \bar E$. But let $B=\{(x,y)|x^2+y^2=1\}\subset \mathbb{R}^2$, and $E=\{(x,y)|x^2+y^2< 1\}$, then $B\subseteq \bar E$, but $B\not\subseteq\overline {E\cap B}=\emptyset$. – user129161 Feb 22 '14 at 15:01
The implication $B\subset\overline{E}\implies B\subset \overline{E\cap B}$ is false, since one can choose $B\cap E=\emptyset$, but $B\subset \overline{E}$. – Xiang Yu Nov 8 '15 at 2:33