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Let $(X,d)$ be a metric space. The sequence $(x_n)$ converge to the set $A\subset X$ (denoted as $x_n\to A$) iff $$ \lim\limits_n d(x_n,A) = 0 $$ where $d(x;A) = \inf\limits_{y\in A}d(x,y)$. Let $\overline A$ be the closure of $A$, and $x_n\to \overline A$. Does it mean that $x_n\to A$?

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Let $\varepsilon>0$, and $n_0$ such that for $n\geq n_0$ $d(x_n,\bar A)\leq \varepsilon$. Let $a_n\in \bar A$ such that $d(x_n,a_n)<2\varepsilon$, and $a_n'\in A$ such that $d(a_n,a_n')\leq\varepsilon$. We have $$\forall n\geq n_0:\quad d(x_n,A)\leq d(x_n,a_n')\leq d(x_n,a_n)+d(a_n,a_n')\leq 3\varepsilon,$$ so $x_n\to A$.

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Thanks! I think now, that we can also show first that $d(x;A) = d(x,\bar A)$ –  Ilya Feb 10 '12 at 10:17
    
Yes, I may be the simplest way. –  Davide Giraudo Feb 10 '12 at 10:18
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