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Is the union of two nowhere dense sets nowhere dense?

Using the following definition:

A nowhere dense set is a subset $E\subset X$ of a metric space (or topological space) $X$ such that $(\overline{E})^o=\emptyset$.

I tried using topological properties like "union of closure of sets is the closure of union", and others. I tried also using the fact that $(\overline{A})^c={(A^c)}^o$ and other complement elementary-set-theory identities.

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What do you mean by can't do a proof using $\dots$? Not allowed to? If you are allowed, the closure of a union of a finite number of sets is the union of the closures. –  André Nicolas Oct 30 '12 at 3:07
    
Can't you prove the closure of the union is the union of the closures? –  Gerry Myerson Oct 30 '12 at 3:07
    
Sorry, edited... –  Gastón Burrull Oct 30 '12 at 3:38
    
Seems, more tricky than expected ...actually, I'm not sure wether this is true though everywhere stated –  Freeze_S Jan 29 at 1:18
    
...though the union of interiors can be strictly smaller than the interior of unions as in $(\mathbb{Q})°\cup(\mathbb{R}\setminus\mathbb{Q})°=\varnothing\subsetneq\mathbb{‌​R}=(\mathbb{Q}\cup\mathbb{R}\setminus\mathbb{Q})°$ –  Freeze_S Jan 29 at 1:34
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1 Answer

up vote 2 down vote accepted

Put another (equivalent) way, a set $A$ is nowhere dense iff for every open set $U$ there is an open set $V$ such that $V\subseteq U$ and $A\cap V=\emptyset$. (I leave it to you to prove the equivalence.) That version should make your task much simpler.

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Ty, Ill try with this definition. –  Gastón Burrull Oct 30 '12 at 3:31
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