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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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What exactly is $A$? –  Nameless Dec 16 '12 at 11:28
    
Corrected, thanks –  user50554 Dec 16 '12 at 11:30
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What if $M=\mathbb R$ with standard metric and the sequence is an enumeration of $\mathbb Q$? –  Hagen von Eitzen Dec 16 '12 at 11:32
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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. –  Grumpy Parsnip Dec 16 '12 at 11:32

2 Answers 2

up vote 5 down vote accepted

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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If the sequence is constant or finally constant then on the contrary, the claim is true. In the first case $\{x_{n}\}_{n=1}^{\infty}=\{x_{n}\}$ for all $n\in\mathbb{N}$, and on the latter case $\{x_{n}\}_{n=1}^{\infty}$ is a finite set. –  Thomas E. Dec 16 '12 at 11:52
    
@ThomasE. That's what I meant by "follows easily" –  Nameless Dec 16 '12 at 11:53
    
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. –  Thomas E. Dec 16 '12 at 11:55

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