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"The number of distinct limits of subsequences of a sequence is finite?"

I've been mulling over this question for a while, and I think it is true, but I can't see how I might prove this formally. Any ideas?

Thanks

3 Answers 3

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Enumerate the rationals as $r_1,r_2,r_3,\dots$. Every real number is the limit of a subsequence of $(r_n)$.

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    Wow, that is fantastic, thankyou.2012-03-15
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No, the following is a counter-example: Let $E: \mathbb N \to \mathbb N^2$ be an enumeration of $\mathbb N^2$, and set $a_n = (E(n))_1$. Then $a_n$ contains a constant sub-sequence $a_{n_i} = k$ for every $k \in \mathbb N$.

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The answer is no. Think for instance about subsequences of $(cos(n))$: after showing that the integers are dense in $[0,2\pi]$, it follows that for every real number $k$ in $[-1,1]$ there exists a subsequence $(cos(n_k))_{k \in \mathbb{N}}$ which converges to $k$. In particular, not only the number of distinct limits of subsequences is infinite, but uncountably infinite.