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How to prove that, for any sequence $(s_n)$ of real number and any real number $z$, the following $2$ statements are equivalent?

$1.$ Every subsequence of $(s_n)$ has a further subsequence that converges to $z$.

$2.$ $(s_n)$ converges to $z$.

So I think we need to prove the "iff" relationship. My process:

"if" :Let $s_{n_{k}}$ be the subsequence of $s_n$, so we know that exits a further subsequence by definition. But to prove that the further subsequence converges, I could not even think of a real example. The only clue I can think of is that we know by theorem that every sequence has a monotonic subsequence, so if we can prove the subsequence is bounded, then we can prove its convergence.

"only if" : If $(s_n)$ converges to $z$, then we know every subsequences converges to the same limit by therorem, then every further subsequence converges to the limit of the subsequence.

I have difficulty translate it to rigorous proof and I have difficulty prove the "if" direction. Could someone help?

CoolKid
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  • For $1\implies 2$, try proving the contrapositive, that $\lnot 2 \implies \lnot 1$. That is, assume that the sequence does not converge to $z$, and use this to construct a subsequence that has no further subsequence that converges to $z$. – florence May 14 '17 at 19:32
  • See this https://math.stackexchange.com/questions/397978/every-subsequence-of-x-n-has-a-further-subsequence-which-converges-to-x-then – Zain Patel May 14 '17 at 19:34
  • Also in your if direction try to the full power of the statement "for every subsequence...". I do not think you are using it currently. – Juanito May 14 '17 at 19:46
  • @florence Would you mind giving me a hint for contrapositive way? – CoolKid May 17 '17 at 02:30
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    Suppose $\lnot 2$. So, if $s_n$ does not converge to $z$, then you can construct a subsequence which does not ever get within $\varepsilon$ of $z$. Now show that no subsequence of this subsequence will converge to $z$. – florence May 17 '17 at 06:06
  • Here's my idea. Suppose $s_n$ doesn't converge to $z$, then $\forall \epsilon, \forall N, \exists n>N$ such that $|s_n-s|\ge\epsilon$. Then we let $n_k=N+1$ for all $k$, then the subsequence diverges as well, then apply for further subsequence. Is it correct? It seems to me like guessing – CoolKid May 17 '17 at 21:22

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