Whats the difference between two set being not nowhere dense and being dense?

Question

I am really confused with the definitions regarding not nowhere dense and dense. I know that for a set $A$ to be dense, I need to show that for any open interval $I \subseteq \mathbb{R}$, $I \cap A \neq \emptyset$

In general talking , for something to be dense, if I were to pick any point in $\mathbb{R}$, that point is either in the dense set or close to that set (the neighborhood of that point has points from that set).

Now I was working on a proof , to show $\mathbb{Q}$ is dense in $\mathbb{R}$ . But then I encounter a question asking to prove Rationals are not nowhere dense,immediately I say this statement is equivalent to rationals are dense in R (same proof as $\mathbb{Q}$ is dense in $\mathbb{R}$) but i guess i am wrong because my teacher didn't approve my proof( ironically) I used the theorem " between any two reals there is a rational " to prove.

We haven't learned about dense sets yet (I am doing some esearch, reading some definitions in our textbook and trying to solve the questiona given) but I didn't find much about nowhere dense and not nowhere dense definitions that I could understand clearly and start my proof. Kindly if anyone could make it clear for me ? A Simple proof could also help

Answer 1

Definition: $X\subseteq\mathbb{R}$ is nowhere dense if the closure $\overline{X}$ of $X$ has empty interior (i.e., $\overline{X}$ does not contain any nonempty open intervals).

1. If $X\subseteq\mathbb{R}$ is dense then it is not nowhere dense.
   Proof: If $X$ is dense then $\overline{X}=\mathbb{R}$.

2. "Not nowhere dense" does not necessarily imply "dense".
   For example, let $X$ be a nonempty open interval. Then $\overline{X}$ contains $I$, so $X$ is not nowhere dense. But also $X$ is not dense unless $X=\mathbb{R}$. For a less trivial example, consider $X=\mathbb{Q}\cap I$ where $I$ is a nonempty open interval. Then $X$ is not nowhere dense, since $\overline{X}$ contains $I$, but $X$ is not dense unless $I=\mathbb{R}$.

3. If $X\subseteq\mathbb{R}$ then the following are equivalent.
   a) $X$ is not nowhere dense.
   b) There is a nonempty open interval $I\subseteq\mathbb{R}$ such that $X\cap I$ is dense in $I$.
   Proof:
   [a to b]. If $X$ is not nowhere dense then $\overline{X}$ contains some nonempty open interval $I$. So any nonempty open sub-interval $J$ in $I$ intersects $X$, which says that $X\cap I$ is dense in $I$.
   [b to a]. If $X\cap I$ is dense in $I$ then $\overline{X}$ contains $I$. So $X$ is not nowhere dense.

Remark: Sets with property 3b are called "somewhere dense". So the full conclusion is that "not nowhere dense" is the same thing as "somewhere dense", which is weaker than "dense" (aka "everywhere dense").

Answer 2

The straightforward definition of $A$ being nowhere dense is $$\operatorname{int}(\overline{A}) = \emptyset\tag{1}$$

i.e. even the closure of $A$ does not contain a non-empty open set. Clearly $\Bbb Q$ has closure $\Bbb R$ (this is what $\Bbb Q$ being dense means) and so the interior of the closure is not empty and so $\Bbb Q$ is not nowhere dense.

The name nowhere dense comes from this alternative definitions/older terminology:

$A \subseteq X$ is called "somewhere dense", if there is a non-empty open subset $O$ such that $O \cap A$ is dense in $O$. And $A$ is called "everywhere dense" if we can take $O=X$. $A$ is called nowhere dense if it is not "somewhere dense".

What we now call just "dense" just to be "everywhere dense". And it's a tautology that a dense (i.e. everywhere dense) set is not nowhere dense. (as "not not" cancels out).

Answer 3

A set is nowhere dense if its closure has empty interior. The rationals being dense in the reals, have as closure the whole real line which contains non empty open sets, hence doesn't have empty interior. Thus, $\mathbb{Q}$ is not nowhere dense (and a way to conclude that is precisely to note that $\overline{Q}=\mathbb{R}$).