Density is one of the ways of measuring how ‘big’ a set is. A set is dense in another set if every subset of the latter set contains a (infinite amount of) point(s) of the former set.
Definition
A set S \subset \mathbb{R} is dense in K \subset \mathbb{R} if every non-empty open set U \subset K contains an element from S.
This means for point in K, there exists a point in S arbitrarily close.
The prime example of a dense set is \mathbb{Q} being dense in \mathbb{R}. Every non-empty open interval of \mathbb{R} contains some rational point p \in \mathbb{Q} by the Archimedean property.
Properties
Denseness has some nice properties.
Every set is dense in itself.
Every set is dense in its closure.
Given three sets A,B,C \subset \mathbb{R}, if A is dense in B and B is dense in C, then A is dense in C.
Nowhere Dense
The opposite of a dense set is one that isn’t dense, anywhere on \mathbb{R}. As we saw in the previous section, every set is dense in itself and in its closure, so we’ll have to specify exactly what we mean.
A set S \subseteq \mathbb{R} is nowhere dense if for every nonempty open interval I, S is not dense in I.
By specifying that S is not dense in a nonempty open interval, we don’t run into problems with S being dense in itself or it’s closure.
A set S \subseteq \mathbb{R} is nowhere dense if \varnothing = \text{int}(\text{cl}(S)).
This theorem is so useful it is often given as the definition of a nowhere dense set. The following are useful properties of nowhere dense sets
Let S be a nowhere dense set. Then every subset of S is also nowhere dense.
Let \{S_i \subset \mathbb{R} \mid i \in I\} be nowhere dense sets for a finite index set I. Then \bigcup_{i \in I} A_i is nowhere dense.
Let I_1 be an arbitrary interval, then there exists some p_1 \in I_1 such that for some \varepsilon_1 > 0, {(p_1 - \varepsilon_1,p_1 + \varepsilon_1)} \cap S_1 = \varnothing since S_1 is nowhere dense.
Next let I_2 = {(p_1 - \varepsilon_1,p_1 + \varepsilon_1)}. Then there exists some p_2 \in I_2 such that for some \varepsilon_2 > 0, {(p_2 - \varepsilon_2,p_2 + \varepsilon_2)} \cap S_2 = \varnothing since S_2 is nowhere dense.
Thus S_1 \cup S_2 is nowhere dense in I_1. This can repeated any finite amount of times.
This argument does not work when applied to a countable union of nowhere dense sets, as the nested interval theorem states that our intersection \bigcap I_n of intervals will contain a point, which may belong to some set S_n.
There also exists the counterexample of \mathbb{Q}, which is a countable union of nowhere dense sets (singletons) but is dense in \mathbb{R}.
The closure of a nowhere dense set is nowhere dense.
This is a simple one liner, since \text{cl}(\text{cl}(S)) = \text{cl}(S) for every set S. By our theorem if S is nowhere dense,
\text{int}(\text{cl}(\text{cl}(S))) = \text{int}(\text{cl}(S)) = \varnothing.
So the closure of S is also nowhere dense.