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İ studied the definition of nowhere dense set and I really can't understand it very well is there simple explanation of the definition and the two following questions. A set  is nowhere dense if  (the closure of  it ) contains no nonempty open ball.

Let $ {(}{X}{,}\mathit{\rho}{)} $ be metric space

1)Show that the complementary of nowhere dense set is every where dense. is the converse is right ? Give an example.

2)on the same metric space .Show that the closure of nowhere dense set is nowhere dense too .

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  • $\begingroup$ You wrote: "A set is nowhere dense if (the closure of it) contains no nonempty open INTERVALS." What is an "interval" in a metric space? Are you sure your question isn't just about nowhere dense subsets of the real line? $\endgroup$ – bof Mar 27 '17 at 8:53
  • $\begingroup$ Part (2) seems easy. A set $A$ is nowhere dense if the closure of $A$ has some property. OK, then the closure of $A$ is nowhere dense of the closure of the closure of $A$ has that property. How is the closure of the closure of $A$ related to the closure of $A$? $\endgroup$ – bof Mar 27 '17 at 8:57
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Just write down the definitions, and write down their negations.

A set $A$ in a metrix space $(X,d)$ is nowhere dense if the closure of $A$ has empty interior, that is, equivalently - its closure does not contain an open ball of the metric space.

A set $B$ is a metric space $(X,d)$ is everywhere dense if for every open set $O\subset X$, the intersection $B\cap O$ is not empty.

Consider now a set $A\subset X$. If the complement is not everywhere dense, then there is some non-empty open set $O\subset X$ such that $O\cap A^c=\emptyset$. It follows that $O\subset A$, and so $A$ has non-empty interior, hence its closure has non-empty interior, hence it is not nowhere dense. It follows that if $A$ is nowhere dense, then $A^c$ is everywhere dense.

As for the converse, the question is whether the complement of a dense set is nowhere dense. This is not true: Take the rational points in $[0,1]$, they are dense. Their complement is the set of irrationals, whose closure is $[0,1]$, so it is not nowhere dense.

The second question was already answered in the comments.

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  • $\begingroup$ Really thanks a lot ....(^__^) $\endgroup$ – user416990 Mar 27 '17 at 9:40

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