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 e > 0, and for all K > 0, we can find an a_n with |a_n-l| > e (n > K), so if we can find an infinite number of such a_ns, 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 e > 0 there can only be finitely many terms a_n satisfying |a_n-l| > e, as the hypotheses dictate.

This is the statement of the convergence of a₁, a₂, ... to l.

Ian