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A friend presented the following proof that sequence converges iff every subsequence converges to the same limit.

Something seems awry, though I didn't know how to assure him.

His proof of the reverse direction was as follows:
Suppose every subsequence converges to the same limit.

Let $J \subset \mathbb{N}$ be such that $\{a_n\}_{n \in J} \to a$. Consider $\mathbb{N} \setminus J$. Then $\{a_n \}_{n \in \mathbb{N} \setminus J} \to a$ by hypothesis. Ergo, $\{a_n\}_{n \in \mathbb{N}} = \{a_n\}_{n \in J} \cup \{a_n \}_{n \in \mathbb{N} \setminus J} \to a$ as well.

Something seems amiss, though I can't quite articulate it.

Anthony Peter
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  • The reserve direction is actually trivial: The sequence itself is a subsequence of itself. Your friend's proof is actually correct and can be made rigorous. –  Sep 15 '15 at 02:13
  • @JohnMa Perhaps it is meant to be a strict subsequence? – Anthony Peter Sep 15 '15 at 02:13
  • Umm.... From your statement I don't see it has to be "strict". Anyway the statement is still correct if you restrict to "strict" subsequences. –  Sep 15 '15 at 02:16
  • @JohnMa Why the long-winded proof here: http://math.stackexchange.com/questions/776899/if-every-convergent-subsequence-converges-to-a-then-so-does-the-original-boun ? – Anthony Peter Sep 15 '15 at 02:19
  • The statement is a bit different. In that question only "convergent subsequence" is given. –  Sep 15 '15 at 02:22
  • Your friend's argument cannot be used in that case, for the subsequence of all $a_n$ with $n\in \mathbb{N}\setminus J$ might conceivably not converge. – André Nicolas Sep 15 '15 at 02:52
  • @AndréNicolas Could you elaborate? – Anthony Peter Sep 15 '15 at 02:53
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    Your condition is that all (presumably infinite) subsequences converge and converge to the same value. The condition in the other problem is that the subsequences that happen to converge all converge to the same value. But the other problem adds the condition of boundedness, which your problem does not ask for. The two problems are very different. – André Nicolas Sep 15 '15 at 02:57

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The problem of the proof of your friend is the lack of clarity about the set $\mathbb{N}/J$. If the set $\mathbb{N}/J$ is infinite then $\{a_n\}_{n\in\mathbb{N}/J}$ is a subsequence of $\{a_n\}_{n\in\mathbb{N}}$ as well as $\{a_n\}_{n\in J}$ and everything goes well. However there is still the (trivial) case that the set $\mathbb{N}/J$ is finite.

corindo
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