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Are there literal interpretations of the terms "nowhere dense" and "dense-in-itself" from which these terms' definitions follow? If I were to guess what it means for a subset $A$ of a topological space $X$ to be "nowhere dense", I'd say it means that for all subsets $B$ of $X$ (these are the "where"s), the closure, in $B$, of $A\cap B$ is not all of $B$.

Question 1: Is this notion equivalent to the true definition of "nowhere dense", which is that $\text{int}(\text{cl}(A)) = \varnothing$? The answer here seems to be lacking according to its sole comment.

Question 2: I can't even think of a literal interpretation of "dense-in-itself". Can anybody offer one which is equivalent, or at least sheds some light on, the true definition, which is that $A$ has no isolated points?

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Unit
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The closure of $A$ in $A$ is all of $A$, so by your definition no space has any nowhere dense subsets. What is true is that $A$ is nowhere dense in $X$ if $\operatorname{cl}_B(A\cap B)\subsetneqq B$ for all non-empty open subsets $B$ of $X$. In very informal terms, you could say that ‘whereness’ requires having non-empty interior: only non-empty open sets take up enough ‘space’ to count.

The closest I can come to giving you an intuitive justification for dense-in-itself is that it’s short for the assertion that every cofinite subset is dense in the set: every subset that is ‘almost’ the whole set (in the sense of cardinality) is dense in the whole set.

Brian M. Scott
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  • It is very satisfactory to me that the only sets with enough "whereness" to be considered a "somewhere" are the non-empty open sets: these are what topology is all about. As for "dense-in-itself", do you mean (to be precise) that for all subsets $B$ of $A$ with $|B| < |A|$ we have $\text{cl}_A(A\setminus B) = A$? – Unit Oct 09 '15 at 20:14
  • @Unit: No, I mean all cofinite subsets: those $B\subseteq A$ such that $A\setminus B$ is finite. Equivalently, you could look only at co-singletons, those $B$ such that $|A\setminus B|=1$. – Brian M. Scott Oct 09 '15 at 20:16
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    Oh, right. And that can be described as follows: $A$ is dense (the plain English meaning) around every one of its points. Great! Thank you! – Unit Oct 09 '15 at 20:31
  • @Unit: You’re welcome! – Brian M. Scott Oct 09 '15 at 20:32