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Suppose $X$ is a metric space, $z \in X$ and $(x_n)$ is a sequence in $X$. Show that if $X$ has a subsequence that converges to $z$, then dist$(z ,$ {$x_n :n ∈ N$}) $= 0$, and show also that the converse need not be true.

I have proved the first part that if $X$ has a subsequence that converges to $z$, then dist$(z ,$ {$x_n :n ∈ N$}) $= 0$.

But I need help in finding the counterexample to show that the converse is not true.

Thank You!!

User8976
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What about taking $X:=[0,1]$, $z:=1$ and $$ x_n:=\begin{cases}1&\text{if }n=1,\\0&\text{otherwise}.\end{cases} $$

Guest
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    More generally, if $z$ itself is one of the elements of the sequence, but is not an accumulation point (that is, there is no subsequence of distinct elements converging to $z$), you have the counterexample you want. – fonini Jan 15 '15 at 12:45