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??
Best Answer
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.