Understanding the definition of nowhere dense sets in Abbott's Understanding Analysis

First of all, I am sorry for asking a question about understanding a definition in a book named Understanding Analysis. But it is my first time to encounter basic topology, so I hope you can excuse me. I have searched previous questions like this and this. I have looked to the wiki page. Still I am having a hard time to understand the following definition:

A set $E$ is nowhere dense if $\overline E$ (the closure of $E$ ) contains no nonempty open intervals.

I am not familiar with other concepts of topology which are not available in the Abbott's Understanding Analysis like balls or interior. I know a set $A$ is dense in $B$ if and only if $\overline A = B$ . For example, $\mathbb Q$ is dense in $\mathbb R$ , because its limit points are all real numbers and its closure gives $\mathbb R$ . Similarly, $\mathbb Z$ is not dense in $\mathbb R$ because it doesn't have limit points and hence its closure is itself. According to my knowledge of denseness, could you help me to understand the above definition with an example?

• real-analysis
• general-topology

$\begingroup$ Nothing to be sorry about! You have searched what you could and asked a good question, and you deserve a good answer. =) $\endgroup$

$\begingroup$ +1 for sorry for asking a question about understanding a definition in a book named Understanding Analyis $\endgroup$

$\begingroup$ Not sure if this helps. $S$ is the Serbs and $B$ is the Bosnians. $B$ is nowhere dense in $S$ if we can take any Bosnian (who isn't also a serb; the analogy breaks a bit here), and put an open fence around him, so that throughout the inside the fence there are only Bosnians. $$$$ On the other hand if there was a place that $B$ was dense in $S$, then there would be some Bosnian who can't fence himself off from Serbs in an open way. That Bosnian would be topologically 'inseperable' in some sense from the Serbs. $$$$ Of course, precisely what `open' means depends on the topology you have. $\endgroup$

$\begingroup$ For arbitrary $S$ and $B$, it may happen that all the Bosnians make fences separating themselves from the Serbs (no matter what size or shape, open or closed, just as long as there are no Serbs). More Bosnians come and stand on top of the fence. Whereas Bosnians previously only roamed inside the fences, now there are also Bosnians directly along all the borders too. (Closure) $$$$ Now if none of the Bosnians find themselves in an open region of only Bosnians, then $B$ is nowhere dense in $S$. In this case the Serbs have contained the Bosnians to isolated points. $\endgroup$

$\begingroup$ @enthdegree Thank you for this Bosnian/Serbian example. It is a nice analogy to the intuition behind the definitions. $\endgroup$

4 Answers 4

Nowhere dense is a strengthening of the condition "not dense" (every nowhere dense set is not dense, but the converse is false). Another definition of nowhere dense that might be helpful in gaining an intuition is that a set $S \subset X$ is nowhere dense set in $X$ if and only if it is not dense in any non-empty open subset of $X$ (with the subset topology).

For example, $\mathbb$ is nowhere dense in $\mathbb$ because it is its own closure, and it does not contain any open intervals (i.e. there is no $(a, b)$ s.t. $(a, b) \subset \mathbb<\bar> = \mathbb$. An example of a set which is not dense, but which fails to be nowhere dense would be $\ \; | \; 0 < x < 1 \>$. Its closure is $[0, 1]$, which contains the open interval $(0, 1)$. Using the alternate definition, you can note that the set is dense in $(0, 1) \subset \mathbb$.

An example of a set which is not closed but is still nowhere dense is $\ \; | \; n \in \mathbb\>$. It has one limit point which is not in the set (namely $0$), but its closure is still nowhere dense because no open intervals fit within $\ \; | \; n \in \mathbb\> \cup \$.

$\begingroup$ I don't want to add yet another answer just to say this, but the nice quantifier-free topological definition of nowhere dense is $\operatorname(\bar S)=\emptyset$. $\endgroup$

$\begingroup$ @MarioCarneiro Good point, this is in fact how I have usually seen it defined in books on general topology (with good reason). I did not use it because the asker said he is not familiar with the concept of an interior. $\endgroup$

$\begingroup$ If you know closure but not interior, another way to write it is $\overline=X$, which can be read "the complement of $\bar S$ is dense". $\endgroup$

$\begingroup$ Your clear answer and examples really helped me to grasp the idea. Thank you very much. $\endgroup$

$\begingroup$ $A$ is dense in $B$ whenever the part of $A$ that lies in $B$, so $A \cap B$ is dense in $B$ in the closure sense. That's the same as the $B \subset \overline$ definition. @HritRoy $\endgroup$

I'm not overly familiar with what's in Abbott's Understanding Analysis, but I'll try to pass from denseness to nowhere denseness using open sets. This is mainly to give another way of looking at these sets.

As you have probably seen, a set $A$ is dense in $\mathbb$ (or, respectively, any topological space $X$) if $\overline = \mathbb$ (respectively, $\overline = X$). An equivalent characterisation of these sets is the following:

Fact: $A$ is dense if and only if $A \cap U \neq \emptyset$ for every nonempty open set $U$.

Being "not dense" would be the opposite of this: "there is a nonempty open set $U$ which is disjoint from $A$". But a "not-dense" set can still be "somewhere dense". For example $A = ( - \infty , -1 ) \cup ( 1 , + \infty )$ is not dense (as it is disjoint from the nonempty open set $(-1.1)$), but every nonempty open set which is not a subset of $(-1,1)$ has nonempty intersection with $A$.

So we can try to strengthen the "not dense" condition. A first attempt would be "has empty intersection with every nonempty open set". Unfortunately this is not a really useful property, as it will only be satisfied by the empty set.

So we can try something in between "not dense" and the "empty" condition above. We would like to say that $A$ has empty intersection with "lots" of open sets. One option is to say that while $A$ may have nonempty intersection with a nonempty open set $U$, we can shrink $U$ to another nonempty open set $V$ which is disjoint from $A$. This turns out to be equivalent to "nowhere denseness".

Let's look at $\mathbb$, $\mathbb$ and the Cantor ternary set $C$ as subsets of $\mathbb$:

• Well, we know that $\mathbb$ is dense in $\mathbb$, so it has nonempty intersection with every nonempty open set. So this set cannot be nowhere dense. (This actually shows that no dense set can be nowhere dense.)
• Suppose that $U$ is a nonempty open subset of $\mathbb$. If $U \cap \mathbb$ is nonempty, but $n$ in the intersection. By definition of openness there is a $\varepsilon > 0$ such that $( n - \varepsilon , n + \varepsilon ) \subseteq U$. Without loss of generality we may assume $\varepsilon \leq 1$. But now $( n , n + \varepsilon )$ is a nonempty open subset of $U$ which is disjoint from $\mathbb$.
• Suppose that $U$ is a nonempty open subset of $\mathbb$ with nonempty intersection with $C$. Picking $x \in C \cap U$ there is a $\varepsilon > 0$ such that $( x - \varepsilon , x + \varepsilon ) \subseteq U$. Pick $n$ so large that $3^ \leq \varepsilon$. As $x \in C$, then $x$ is an element of the $n$-th set in the usual construction of $C$ (by removing the middle thirds). Recall that at this stage is the union of disjoint closed intervals of length $3^$ (yeah, I'll start with the $0$th stage being $[0,1]$). It follows that the closed interval, call it $[a,b]$, at this stage containing $x$ is a subset of $( x - \varepsilon , x + \varepsilon ) \subseteq U$. But then the open middle third of this closed interval $( a + \frac3 , b - \frac3 )$ is removed at the next stage of the iterated construction of $C$, and is therefore disjoint from $C$, and it is also a subset of $U$.

(Admittedly, using the text-book definition of nowhere denseness makes it somewhat easier to show that $\mathbb$ and $C$ are nowhere dense, but as I said above, I'm just trying to provide another way at looking at this property.)