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Let $(M,d)$ be a metric space and $\{x_n\}_{n=1}^\infty\subset M$ be a sequence. Prove that $\forall n\in\mathbb N,\quad\exists \varepsilon> 0 \;B(x_n,\varepsilon)\cap \{x_n\}_{n=1}^\infty = \{x_n\}$

Any hint? I don't know how to start this proof. Maybe reductio ad absurdum? Counterexamples?

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    Hint:Consider a sequence that limits to a certain value and then add that value as the first term of the sequence.2012-12-16

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Consider the sequence $ x_n = \begin{cases} \frac{1}{k} & n=2k \\ 0 & n=2k+1 \end{cases}$

The idea is that if $x_n \to \ell$ and there exists $k$ such that $x_k=\ell$ then the property you say is definitely false.

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This is not generally true.

If the sequence is constant or finally constant your claim is true and follows easily.

If this sequence is non constant for $n\ge N$ then

1) $x_n\nrightarrow x_N$. Thus, $\exists \epsilon>0\forall k\in \mathbb{N}\ n\ge k\Rightarrow \left|x_n-x_N\right|>\epsilon$ and...

2)$x_n\to x_N$. Then you can't continue. Such sequences exist as pointed out by Beni Bogosel.

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    Now after the edit I see what you meant. It seemed first that the proposition being generally not true follows easily by considering a sequence that is constant or finally constant. Thanks for clarifying it.2012-12-16