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(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!

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  • $\begingroup$ You're right of course! I made a typo and will edit the question. $\endgroup$ – Étienne Bézout Mar 24 '14 at 11:02
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The set of rationals on the real line has this property (so do the irrationals, for that matter). The rationals have empty interior, but their closure is the whole real line with obviously non-empty interior.

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.

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  • $\begingroup$ @bof: Thanks for pointing that out! $\endgroup$ – Étienne Bézout Mar 24 '14 at 16:35
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Of course it can happen...consider $\mathbb{Q}$ and its closure $\mathbb{R}$.

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  • $\begingroup$ Thanks! However, I accepted Arthur's answer because it was a bit more elaborate. $\endgroup$ – Étienne Bézout Mar 24 '14 at 11:05
  • $\begingroup$ it's ok as long as you understand the concepts :) $\endgroup$ – wanderer Mar 24 '14 at 11:51

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