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In a metric space X, let A be a subset of X. Defining a set A to be nowhere dense if the interior of its closure is empty.
Proposition: A is nowhere dense if and only if for each non-empty open set on X there is an non-empty open subset disjoint from the closure of A.
How can i prove it? I tried to do both direction, using logical implications and also tried to use reduction to absurd, but i was not successful.
HINT: Suppose that $A$ is nowhere dense. You want to show that if $U$ is a non-empty open set in $X$, then there is a non-empty open set $V\subseteq U\setminus A$. You know that $\operatorname{cl}A$ does not contain any non-empty open set, so subtracting it from an open set shouldn’t remove very much. You know that if $W$ is open and $C$ is closed, then $W\setminus C$ is open, so $U\setminus\operatorname{cl}A$ is open. Does it work for $V$?
Now suppose that every non-empty open subset $U$ of $X$ has a non-empty open subset disjoint from $A$; you want to show that $\operatorname{cl}A$ has empty interior. That is, you want to show that $\operatorname{cl}A$ does not contain any non-empty open set. Suppose, on the contrary, that $\varnothing\ne U\subseteq\operatorname{cl}A$, where $U$ is open. Apply your hypothesis to $U$ to get an immediate contradiction.
