Can $\overline{Y}$ have non-empty interior if $Y$ has empty interior

(below, $\overline{Y}$ denotes the closure of $Y$)

Given a metric space $X$ let us define a subset $Y$ to be nowhere-dense if and only if $\overline{Y}$ has empty interior. It is obvious that if $\overline{Y}$ has empty interior, then so does $Y$, being a subset of $\overline{Y}$. But is it possible that $Y$ has empty interior and yet $\overline{Y}$ has not? How would such a set "look"?

Thanks!

• You're right of course! I made a typo and will edit the question. Mar 24 '14 at 11:02

Nowhere dense means exactly that this is not the case. The rationals are everywhere dense, and if you take some subinterval, like $[0, 1] \cup \Bbb Q$, then it's dense on that interval. A set that is nowhere dense on the real line might be a discrete set of points, like the integers.
Of course it can happen...consider $\mathbb{Q}$ and its closure $\mathbb{R}$.