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I'll state the question first.

In any metric space $(X,d)$, assume that $(x_n)$ is a sequence such that $x_n \to x$ for some $x \in X$. If $(y_m)$ is a sequence in $\{x_n: n \in \mathbb{N}\} \cup \{x\}$ such that $y_m \to x$ then is it true that $(y_m)$ must eventually be a subsequence of $(x_n)$?

My definition for an "eventually subsequence" is : there is a map $k:\mathbb{N} \to \mathbb{N}$ such that for all $m \in \mathbb{N}$ there is $n \in \mathbb{N}$ such that $y_m=x_{k(n)}$ and $k$ has the property that there should exist $M \in \mathbb{N}$ such that for all $n_1,n_2 \geq M$ if $n_1<n_2$ then $k(n_1) < k(n_2)$.

I guess this is intuitively a reasonable claim. But I can't prove it rigorously. Any help (counter-examples, proofs etc.) will be appreciated. Thanks and regards.

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What if a term of $(x_n)$ were equal to $x$? (Note you could also take $y_n=x$ for all $n$; this would not be a subsequence of $(x_n)$ unless infinitely many $x_n$ were equal to $x$.) –  David Mitra Mar 28 '12 at 18:12
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What if $y_m = x$ for all $m$? –  TMM Mar 28 '12 at 18:14
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Or - if all the elements of the series are different - swap the even-numbered elements of the series with adjacent odd numbered ones ($x_{2n}$ with $x_{2n-1}$) –  Mark Bennet Mar 28 '12 at 18:24
    
+1 I think it is a good question that is worth to think about. –  AD. Mar 28 '12 at 19:20

2 Answers 2

up vote 4 down vote accepted

Clearly wrong. Consider $x_n=\frac1n$, $y_{2n-1}=\frac1{n+1}$ and $y_{2n}=\frac1n$ for every $n\geqslant1$, then $k(2n)=n\lt n+1=k(2n-1)$.

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I assumed your definition was a little mixed up and that you really wanted to assume that for every $n$ there exists $k(n)$ such that $y_n=x_{k(n)}$. –  Did Mar 28 '12 at 18:23
    
Yes my definition was a mess. Thanks for the answer. –  Sayantan Mar 29 '12 at 0:55

I think the following is a counter-example (if I have understood you correctly). Let $$a_n:=\frac1n\\b_{2n}:=\frac{1}{2n-1}\\b_{2n-1}:=\frac{1}{2n}$$ Note that $\lbrace a_n| \textrm{ }n\in\mathbb{N}\rbrace = \lbrace b_n| \textrm{ }n\in\mathbb{N}\rbrace$ but $(b_n)_n$ is not a subsequence of $(a_n)_n$ from any place on.

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