# Question regarding nowhere dense and everywhere dense sets

I was just refreshing my topology basics and started wondering about this.I do hope my question is mathematically sound. Anyway my question is whether every topological space has non-trivial dense sets or nowhere dense sets or for that matter even separable sets?? Does the presence of such sets affect the geometry of a space in any manner?? Does this question make sense in the first place??I wasn't able to get a clear picture.

Also as an exercise, I was trying to prove that if $A$ is nowhere dense in a metric space $(X,d)$ then this is equivalent to saying that every non-empty open set in $X$ has a non-empty open subset disjoint from $A$.

I was thinking along these lines if suppose $\exists \ U$ open in $X$ such that $\forall \ V$ open in $U$, $V \cap U \neq \phi$ . Then this means for every $x$ and some $\epsilon$ such that $B(\epsilon ,x) \subset U$ we have $B(\epsilon ,x) \cap A \neq \phi$ . Now if $x_1$ belongs to this intersection, then I can find $B(\epsilon_1, x_1) \subset B(\epsilon ,x)$ and further $B(\epsilon_1, x_1) \cap A \neq \phi$ and this can go on. Ultimately I feel that this shows that there has to be an open ball $B(\epsilon_n , x_n)$ contained in $A$. Is this line of thinking going to work??Or is there any other way??

The answer to your first two questions is no:

In a discrete space (every subset is open) there are neither any proper dense subsets nor any nowhere-dense subsets. (If the space has at least two points.)

As for the second question, I don't think you need that X is a metric space.

Let $A$ be nowhere-dense and let $U$ be a non-empty open subset of $X$. If $A\bigcap U$ is empty, we're done. If not, $\bar{A}\bigcap U$ is non-empty. By hypothesis, $U$ is not contained in $\bar{A}$, so $U-\bar{A}$ is non-empty, and is therefore the desired open subset.

For the other direction, suppose every non-empty open subset has an open subset disjoint from $A$. If $\bar{A}$ has non-empty interior, that would be an open set with no open subset disjoint from $A$, a contradiction. Thus $\bar{A}$ has empty interior, and $A$ is nowhere-dense.

• Thank you. I am still curious as to how the geometry of space is influenced by such sets though. – Vishesh Oct 24 '12 at 7:19
• If you could explain what you mean by "geometry", I might be able to help. In general, topological spaces can be quite pathological, with little "geometric" intuition underlying their behavior. – user45071 Oct 25 '12 at 14:56
• Well I guess geometry wouldn't make much sense as most geometrical notions like compactness,etc are studied independently on their own.I guess what I was thinking about would be something like whether such a space has "holes"?? or do the surfaces in these spaces have irregularities that wouldn't otherwise be seen in other spaces.I know this seems like hand waving. I am not very clear myself. Its juts a vague notion in my head.I have been recently introduced to manifolds and all, so in a way I was looking at this in that context. – Vishesh Oct 25 '12 at 15:30
• Well, manifolds are certainly going to be separable and have nowhere dense subsets, since they are realizable as subsets of Euclidean space of sufficiently high dimension. (The Whitney Embedding Theorem.) If you want to talk about "holes" you should be looking at notions of connectedness and the fundamental group. Dense and nowhere dense sets don't have much to do with "holes" in the geometric sense. – user45071 Oct 25 '12 at 17:55
• Thanks a lot. I am not familiar with Whitney's embedding theorem or any algebraic topology for that matter. But I will take your word for it.I think with my current background, I cant expect to probe too deep. – Vishesh Oct 26 '12 at 14:02