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What is the practical meaning of a subset being nowhere dense on another set? What does it mean apart from the definition (its closure to have no interior points)?

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Historically, I think the notion arose as a natural opposite condition to that of being everywhere dense. Suppose that $D$ is everywhere dense in $\mathbb R$. Then $D$ is also dense in every open interval of $\mathbb R$. Conversely, if $D$ is dense in every open interval of $\mathbb R$, then $D$ is everywhere dense in $\mathbb R$. The converse is immediate because $\mathbb R$ itself is an open interval, but if this seems like cheating, note that the result still holds if we restrict ourselves to bounded open intervals. Thus, we can say that $D$ is everywhere dense in $\mathbb R$ if and only if $D$ is "everywhere locally dense in $\mathbb R$".

Therefore, a subset $E$ of $\mathbb R$ is not everywhere dense in $\mathbb R$ if and only if $E$ is not dense in some open interval (i.e. $E$ is not locally dense somewhere).

A really strong way for a subset $N$ of $\mathbb R$ to not be everywhere dense in $\mathbb R$ is the property of not being dense in every open interval (i.e. $N$ is not locally dense everywhere). This stronger way is exactly what being nowhere dense means.

It might help to think of these properties as describing sets that are everywhere big, somewhere big, and nowhere big. Or, in place of "big", you can use "thick".

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  • $\begingroup$ Is it true that whenever a subset is nowhere dense this can in fact mean that a small perturbation of its elements will violate its properties? If so in what way can this be done? It is an immediate consequence of the non density? $\endgroup$ – Catherine Mar 28 '13 at 15:41
  • $\begingroup$ @Catherine: You'll probably have to define what a "small perturbation" is to get anywhere with this line of thinking. For example, one might say that two compact sets are small perturbations of each other if their Hausdorff distance is small. However, using this interpretation every compact set is arbitrarily close to a finite set of points, and every finite set of points is nowhere dense in a very strong way. On the other hand, if we're talking about small (or large) translations, then these density properties don't change. $\endgroup$ – Dave L. Renfro Mar 28 '13 at 16:10
  • $\begingroup$ I will be more specific about my problem. Consider the following set F={A: 3-by-3 complex matrices that satisfy a property P}. Proving that F is nowhere dense does it lead to the fact that if we consider a small perturbation of the matrix A\in F, i.e. just translate its non diagonal entries by a small \epsilon>0, then the new matrix will not satisfy property P? $\endgroup$ – Catherine Mar 28 '13 at 17:43
  • $\begingroup$ I think this can fail in a major way. There's an old result of Scheeffer's from 1884 that says if $P$ is a perfect nowhere dense set in $\mathbb R$ and $Z$ is any countable subset of $\mathbb R$ (for what follows, assume $Z \subseteq P$), then for a dense set of $b \in {\mathbb R}$ (in particular, for arbitrarily small values of $b$) the $b$-translate of $P$ will have empty intersection with $Z$. Surely an analogous result holds in ${\mathbb R}^{18}$ (your setting), at least for arbitrarily small translations. $\endgroup$ – Dave L. Renfro Mar 28 '13 at 18:46
  • $\begingroup$ I will look into it! I think you are right! Thank you very much for your response to my problem! $\endgroup$ – Catherine Mar 28 '13 at 19:06

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