A set $A$ in a metric spaces $(E, d)$ is nowhere dense if $(\overline{A})^c$ is dense, i.e, if $\,\,\,\overline{(\overline{A})^c}=E$.
Let $(a_n)$ a sequence of points of $E$ that converges to $a\in E$.
Is the set $A=(a_n)$ nowhere dense??
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Not necessarily: what if $E=A\cup\{a\}$, and the points of $A$ are all distinct? For example, take $$E=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$$ and $a_n=\frac1n$ for $n\in\Bbb Z^+$. Then $A$ is actually a dense open subset of $E$.
Brian M. Scott
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Brian $A$ is open? – Roiner Segura Cubero May 01 '13 at 16:59
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@Roiner: $A=\left{\frac1n:n\in\Bbb Z^+\right}$ is indeed open in $E$: it consists entirely of isolated points. – Brian M. Scott May 01 '13 at 17:01