Edwin Koh

Posted on Thursday, 20 February, 2003 - 08:33 pm:

Suppose a sequence of real numbers {a_n } has the property that every subsequence has a convergent (sub)subsequence and all these convergent (sub)subsequences have the same limit. Why is {a_n } necessarily convergent?

George Walker

Posted on Thursday, 20 February, 2003 - 11:36 pm:

Surely the sequence {an} itself is a subsequence, thus converges to a limit l, so every other subsequence must converge to this limit (or a refinement of this argument, I'm not feeling very analytical this evening)

Kerwin Hui

Posted on Friday, 21 February, 2003 - 01:56 am:

George,

{a_{n}}

is not necessarily the sub-subsequence.

If ${a_{n}}$ has the required property (any subsequence has a subsubsequence converging to $l$ ), and suppose we don't have convergence. This gives for some $ε > 0$ , and for all $K > 0$ , we can find an $a_{n}$ with $| a_{n} - l | > ε$ ( $n > K$ ), so if we can find an infinite number of such $a_{n}$ s, and these form a subsequence of ${a_{n}}$ , which cannot have a subsubsequence converging to $l$ .

Kerwin

Edwin Koh

Posted on Friday, 21 February, 2003 - 08:07 am:

Thanks, Kerwin. I wonder if it's possible to avoid a proof by contradiction.

Ian Short

Posted on Friday, 21 February, 2003 - 11:20 am:

Proofs by contradiction can often be avoided by suitably reformulating the argument. Not that it necessarily makes things clearer, as is the case here. Briefly,

given any $ε > 0$ there can only be finitely many terms $a_{n}$ satisfying $| a_{n} - l | > ε$ , as the hypotheses dictate.

This is the statement of the convergence of $a_{1}$ , $a_{2}$ , ... to $l$ .

Ian