9
$\begingroup$

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?

  • 1
    For the final part: it means that the (relative) closure of $E \cap B$ (here $B$ is a ball) inside $B$ is all of $B$.2011-08-17
  • 1
    You may want to read about the [Cantor Set](http://en.wikipedia.org/wiki/Cantor_set) for an interesting non-trivial example.2011-08-17
  • 2
    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.2011-08-17
  • 0
    @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.2011-08-17
  • 0
    @Theo: Hmm, one can then control the number of balls in which the set is dense by slightly modifying your example ${\mathbb Q}\cap\bigcup_{n\in{\mathbb Z}}[2n,2n+1]$.2011-08-17
  • 0
    It means that $\overline{E}$ contains a ball. Equivalently, $E$ is nowhere dense if and only if the interior of $\overline{E}$ is empty.2011-08-17
  • 0
    Yes, essentially you can freely choose the intervals in which your set is dense by just taking the rationals inside those intervals and throwing the others out. Maybe this modification is also revealing: $\left(\mathbb{Q} \cap \bigcup_{n \in\mathbb{Z}} [1000n, 1000n + 1] \right) \cup \frac{1}{100000000} \cdot \mathbb{Z}$. Can you see in which intervals this set is dense?2011-08-17
  • 2
    @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.2011-08-18
  • 0
    @Jack,@Theo: It’s potentially confusing to use ‘any’ when what is meant is ‘every’ or ‘each’. ‘What is an example of a set $E$ that is dense in every ball?’ and ‘$\mathbb{Q}$ is dense in every ball of $\mathbb{R}$’ are less likely to be misinterpreted or cause the reader to hesitate.2011-08-18
  • 0
    @Brian: thanks... In principle I know but I keep making this slip.2011-08-18

1 Answers 1

6

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.

  • 2
    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 $B\not\subseteq\overline {E\cap B}=\emptyset$.2014-02-22
  • 0
    The implication $B\subset\overline{E}\implies B\subset \overline{E\cap B}$ is false, since one can choose $B\cap E=\emptyset $, but $B\subset \overline{E}$.2015-11-08