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?

4 Answers

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, $\mathbb{Z}$ is nowhere dense in $\mathbb{R}$ because it is its own closure, and it does not contain any open intervals (i.e. there is no $(a, b)$ s.t. $(a, b) \subset \mathbb{\bar{Z}} = \mathbb{Z}$. An example of a set which is not dense, but which fails to be nowhere dense would be $\{x \in \mathbb{Q} \; | \; 0 < x < 1 \}$. 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 $(0, 1) \subset \mathbb{R}$.

An example of a set which is not closed but is still nowhere dense is $\{\frac{1}{n} \; | \; n \in \mathbb{N}\}$. 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 $\{\frac{1}{n} \; | \; n \in \mathbb{N}\} \cup \{0\}$.

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 $\mathbb{R}$ (or, respectively, any topological space $X$) if $\overline{A} = \mathbb{R}$ (respectively, $\overline{A} = X$). 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 $\mathbb{Q}$, $\mathbb{Z}$ and the Cantor ternary set $C$ as subsets of $\mathbb{R}$:

• Well, we know that $\mathbb{Q}$ is dense in $\mathbb{R}$, 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 $\mathbb{R}$. If $U \cap \mathbb{Z}$ 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 $\mathbb{Z}$.

• Suppose that $U$ is a nonempty open subset of $\mathbb{R}$ 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 $3^{-n} \leq \varepsilon$. 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 $3^{-n}$ (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 $( a + \frac{b-a}3 , b - \frac{b-a}3 )$ 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 $\mathbb{Z}$ and $C$ are nowhere dense, but as I said above, I'm just trying to provide another way at looking at this property.)

Very good explanations in the two answers above, but putting things in the simple setting below seems to better reflect the idea behind the notion of a nowhere dense set. Despite the fact that the original question insists on giving an answer without using the notions of an interval or an open set!, I think the very question of "How can we have a more comfortable understanding of a nowhere dense set?" might be of interest for a broader set of audience.

Just recalling that an open set is simply an element of a topology on a set $X$. For example in the real numbers, open sets exclusively consist of the empty set, all intervals and their arbitrary unions; this obviously includes the set of real numbers as an open set.

In a topological space $X$ (those who are unfamiliar with the notion of topology, can simply think of the set of real numbers together with all intervals and their arbitrary unions as mentioned earlier):

1. A subset $Y \subseteq X$ is called to be dense when we have $\overline{Y}=X.$

A set might not be dense, but it might be the case that it is dense locally according to a suitable part of the whole space; so one can consider the following definition:

2. 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.

In fact, a subset is somewhere dense if some part of its closure generates a non-empty open set.

3. 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.

Putting aside the trivial cases, it is interesting to note that if $X$ is Hausdorff and contains a dense subset $Y,$ then by taking any open subset $U\subseteq X$ we can show that the closure of $Y\cap U$ equals the closure of $U.$ Hence, we can build lots of non-dense somewhere dense sets in $X.$

Well , you may think density (or non density) as a relative characteristic of a set where no-where density is a individual characteristic.

Say for example , $\mathbb{Q}$ is dense to $\mathbb{R}$ but not to itself where $\mathbb{Z}$ is dense to itself but not to any other set that embeds it. Thus density depends on how you embeds a set .

On the other hand you may see that no matter which set embeds $\mathbb{Z}$ , it remains no where dense and the direct opposite holds for $\mathbb{Q}$ .

To better understand the scenario you may use an alternative but equivalent definition of no where dense sets i.e. " A set G will be called no-where dense if closure of G will have empty interior i.e. $G^{closure}_{int}$ = $\phi$ ."

Now , as $G^{closure}$ is a closed set, if $\delta G ≠ \phi$ every limit point of $G$ except the boundary points are interior points . Thus the above definition implies either $\delta G= \phi$ or for all $ g \in G^{closure} , g\in \delta G$ .

Now to get the taste of equivalence observe that if a set doesn't have a interior point it can't have an nonempty open interval I $\subset G$ as if if the length of the interval is l , we may take a point $i \in I$ and make a ball $B^{i}_{\epsilon }$ , where $\epsilon < \frac{l}{2}$ , such that $B^{i}_{\epsilon} \subset I \subset G $ . But then i becomes an interior point of G which is contradictory to our assumption .

Now you may see the examples given above through it . $\mathbb{Z}$ is no where dense as it doesn't have any limit point . What for the sequence , $S=\{\frac{1}{n} : n\in\mathbb{N}\}$ ? It has the limit point 0 but 0 is it's boundary point [ every n.b.d of 0 will contain negetive numbers which will not be in the sequence ] ,so the interior remain empty .

What for cantor set ? For cantor set we may use a constructive argument . Let , $I= [0,1]= [0, \frac{1}{3}]\cup[\frac{1}{3} , \frac{2}{3}]\cup[\frac{2}{3}, 1]$

Let now , $ I_{0}= [\frac{1}{3} , \frac{2}{3}]$

$[0,\frac{1}{3}]= [0,\frac{1}{9}]\cup[\frac{1}{9},\frac{2}{9}]\cup[\frac{2}{9},\frac{1}{3}]$

$[\frac{2}{3},\frac{7}{9}]\cup[\frac{7}{9} , \frac{8}{9}]\cup[\frac{8}{9}, 1]$

Let $I_{1}=[\frac{1}{9} , \frac{2}{9}]\cup[\frac{7}{9}, \frac{8}{9}]$

Let d(a,b)= |b-a| a metric on $\mathbb{R}$ . Thus , $d(I_{1})=\frac{1}{3}$ and $d(I_{2})=\frac{2}{9}$ .

Continuing in the fashion it can be seen that , $d(I_{n})= \frac{2^n}{3^n+1}$ .

If we let , $S_n= \sum_{i=0}^{n} d(I_{n})$, as $n\rightarrow \infty$ , $S_{n}\rightarrow 1$ .

Thus the part excluded becomes the whole interval , accord to your definition there can't be any open interval the set can contain , thus it becomes no where dense .