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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

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2 Answers 2

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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What's nice about this example is that not only the number of limits is infinite, it's strictly larger than the size of the sequence! :-) –  Asaf Karagila Mar 15 '12 at 18:56
    
Wow, that is fantastic, thankyou. –  Solaris Mar 15 '12 at 22:19

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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