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

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.)