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Can you give me a hint, why the following characterisation of a nowhere dense set $A$ (where my definition of n.d. is $ \textrm{Int(cl}(A))=\emptyset$) should hold:

For all nonempty open set $U$ there exists an open set $V$, such that $V \subseteq U$ and $A\cap V = \emptyset$ (I found this characterisation on the French Wikipedia article for n.d. sets).

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I think you meant $A \cap V = \emptyset$. –  sdcvvc Jan 3 '12 at 16:48
    
Oh yes, thanks. –  temo Jan 4 '12 at 13:00
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temo: the condition is still not what I meant, it's $A$ not $U$ –  sdcvvc Jan 4 '12 at 13:15
    
From just the term "Nowhere dense", the latter classification sounds more like it should be the "real" one. But that's just me rambling. –  Arthur Jan 4 '12 at 15:35

1 Answer 1

up vote 6 down vote accepted

(a) $\operatorname{Int}\overline A=\emptyset$ $\Leftrightarrow$
(b) There is no non-empty open set $U\subseteq\overline A$. $\Leftrightarrow$
(c) If $U$ is a non-empty open set then $U\setminus\overline A\ne\emptyset$. $\Leftrightarrow$
(d) If $U$ is a non-empty open set then there exists a non-empty open open set $V$ such that $V\subseteq U\setminus\overline A$. $\Leftrightarrow$
(e) If $U$ is a non-empty open set then there exists a non-empty open open set $V$ such that $V\subseteq U\setminus A$.

Equivalence of the last two conditions follows from $\operatorname{Int} (U\setminus A)=U\setminus\overline{A}$.

  • The inclusion $\operatorname{Int} (U\setminus A)\supseteq U\setminus\overline{A}$ is clear, since $U\setminus\overline{A}$ is an open set and $U\setminus\overline{A}\subseteq U\setminus A$.

  • On the other hand, if $V\subseteq U\setminus A$ is an open set, then $V\cap A=\emptyset$. This implies $V\cap\overline A=\emptyset$. We have shown that $V\subseteq U\setminus \overline A$ holds for every open subset $V$ of $U\setminus A$. This means that $\operatorname{Int} (U\setminus A)\subseteq U\setminus\overline{A}$.


You can also find similar proof in the book Elements of Metric Spaces by M.N. Mukherjee p.89. (I found this book simply by typing "let A be nowhere dense" into google, I do not know how good this book is. However, proof of this result seems fine.)

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