# What is the 'meaning' of nowhere dense set?

In some books, nowhere dense set is defined to be $int(\bar A)=\emptyset$ but meanwhile is defined to be $int(A)=\emptyset$ in some books(e.g. Munkres).

So what is the 'meaning' (i.e motivation, intuitive/geometric meaning etc.) of nowhere dense set? Thank you.

• I think it should be $\text{int}(\overline{A})=\emptyset$ but not $\text{int}(A)=\emptyset$. For example, rational numbers are dense but $\text{int}(\mathbb{Q})=\emptyset$. – velut luna May 11 '16 at 14:57
• So far as I can see, Munkres doesn’t define nowhere dense at all; he just talks about closed sets with empty interior. In any case, the correct definition is that $\operatorname{int}\operatorname{cl}A=\varnothing$. – Brian M. Scott May 11 '16 at 15:11
• Munkres definition is correct for closed sets $A$. Perhaps he only uses the term "nowhere dense" for closed sets. Or (as Brian says) perhaps he never uses the term at tll. – GEdgar May 11 '16 at 22:01

You can think of the word "nowhere" as meaning "in no open set". So a subset $A \subset X$ is nowhere dense if it is not dense in any open set, or more precisely: for each open subset $U \subset X$ the set $A \cap U$ is not a dense subset of $U$ (with respect to the subspace topology on $U$).
A set $A \subseteq X$ is dense in $X$ if every element of $X$ is either in $A$ or a limit point of $A$. Hence the meaning of $A$ being nowhere dense is for any $x\in X$, there exists an open set $V$ containing $x$ such that $(V-\{x\})\cap A=\emptyset$. In a metric space, roughly speaking, this means at everywhere it is possible to choose a sufficiently small region so that it contains at most one point of $A$.
• FYI, a formulation of "nowhere dense" in metric spaces that I have found useful is the following: Let $X$ be a metric space and let $E \subseteq X.$ We say that $E$ is nowhere dense in $X$ if (and only if) every nonempty open ball in $X$ contains (in the subset sense) a nonempty open ball that is disjoint from $E.$ – Dave L. Renfro May 11 '16 at 17:00
• @velutluna This answer is not correct! For example, $A = \{0\} \cup \{\frac1n \mid n \in \mathbb{N} \}$ is nowhere dense, yet every neighbourhood of $0$ contains infinitely many elements of $A$. (Note that the correct ‘version’ of this type of intuition is given in Dave L. Renro‘s comment.) – John Don Feb 2 '19 at 14:50
A set $A$ is nowhere dense if every nonempty open set contains a nonempty open set which is disjoint from $A.$