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Recently I came across the fact that the intersection of two dense sets is a dense set.

Take $\Bbb{R}$. Clearly, both $\Bbb{Q}$ and $\Bbb{R}\setminus\Bbb{Q}$ are dense in $\Bbb{R}$. However, $\Bbb{Q}\cap (\Bbb{R}\setminus\Bbb{Q})=\emptyset$. How can the $\emptyset$ be dense?

freebird
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    who says the intersection of dense subsets is dense? – Ittay Weiss Jun 17 '14 at 10:49
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    as you see, it is not true. – mesel Jun 17 '14 at 10:50
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    This is a good counterexample to the false statement that the intersection of dense subsets is dense. – egreg Jun 17 '14 at 10:51
  • @IttayWeiss- I have to prove that the union of two nowhere dense sets is nowhere dense. Let $A$ and $B$ be two such nowhere dense sets. Then $A'$ and $B'$ are clearly dense in $X$. If I can prove that $A'\cap B'$ is dense, then this is equivalent to saying $(A'\cap B')'=A\cup B$ is nowhere dense. – freebird Jun 17 '14 at 10:52
  • that is not the correct way to go about it. Look closer at the definition of nowhere dense sets. – Ittay Weiss Jun 17 '14 at 10:55
  • @IttayWeiss- The complement of a nowhere dense set may not be dense. Is this correct? – freebird Jun 17 '14 at 11:04
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    @freebird The complement of a nowhere dense set is always dense - indeed, one definition of a nowhere dense is that the complement of its closure is dense. But the complement of a dense set is not necessarily nowhere dense, as in the example in your question. So you while you can't prove that the intersection of two arbitrary dense sets is dense (because this is false!) the sets $A'$ and $B'$ in your setup have the stronger property that their complements $A$ and $B$ are nowhere dense. I agree with Ittay that this is probably not the best way to think about the problem though. – mdp Jun 17 '14 at 11:23
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    @freebird: You haven't accepted any of the answers to your questions. You may want to take a look at this. – Michael Albanese Jun 17 '14 at 15:40

1 Answers1

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Result: The intersection of two open dense subsets $U,U'\subset X$ of a topological space $X$ is dense.

To prove this, it is best to define "dense" as: " has non-empty intersection with an arbitrary non empty open subset $ \emptyset \subsetneq V\subset X$".
So, in our case, $V\cap U\neq \emptyset $ and then $(V\cap U)\cap U'\neq\emptyset$ also since $V\cap U$ is a non empty open subset of $ X$ too.
Thus $(V\cap U)\cap U'=V\cap( U\cap U')\neq\emptyset$, proving (by our definition) that $ U\cap U'$ is dense.

Georges Elencwajg
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