Claim:
$Y$ is nowhere dense $\iff$ $X\setminus \overline{Y}$ is dense.
Proof
$Y$ is nowhere dense $\iff$ $Int(\overline{Y})=\emptyset$
$\iff$ $\forall y\in\overline{Y}, \forall r>0, \quad B_r(y)\cap (X\setminus \overline{Y})\neq \emptyset$ (by definition of interior)
$\iff$ $\forall y\in Y$, $y$ is adherent to $X\setminus \overline{Y}$.
$\iff$ $\overline{X\setminus \overline{Y}}=X$
I am just beginning to study metric spaces and still getting used to proofs. It feels to me like everything is going alright until the last equivalence. I am not sure what happens if $y$ is an element of the boundary of $Y$. If $y$ is in the boundary of $Y$, then would that be included by the closure of $X\setminus \overline{Y}$? Any suggestions are welcome. Thanks!
Best Answer
This looks good to me.
The boundary of $Y$ is given by $\partial Y =Y \setminus Int(Y)$. But of course, $Int(Y) \subset Int(\bar{Y})$ and so $Int(Y) = \emptyset$. That means that the boundary of $Y$ is the whole of $Y$. That is, $\partial Y = Y$.
As you showed, $X \setminus \bar{Y}$ contains $Y$ in its closure, hence it contains $\partial Y$.