Lecture 9: Limsup, Liminf, and the Bolzano-Weierstrass Theorem

MIT 18.100A Real Analysis, Fall 2020
Instructor: Dr. Casey Rodriguez

Does a bounded sequence have a convergent subsequence? We introduce limit inferiors and limit superiors to prove this is the case (known as the Bolzano-Weierstrass theorem).

License: Creative Commons BY-NC-SA

[nicolasg.b.]

Hey self-learners, be aware of a minor typo in the example of x_n = (-1)^{n} @ 40:11 It should be k >= n in both sup and inf. Not n >= k.

[georgeyang]

51:19 it is not straightforward to me why the BW theorem is correct given the previous theorem.

[gabbiewolf]

1:02:40 Alternatively instead of defining an n_0 is one can define the sequence of indices for the lower and upper bounding sequences as
n'_k =
{1; k = 1
{n_(k-1)+1; k>1
That way you simply have
a_(n'_k) - 1/k < x_(n_k) <= a_(n'_k)

[gabbiewolf]

17:55 There's a problem with the proof here. x_n < sqrt(2/[n-1]) only holds for n > 1. For n = 1 the right hand side is undefined. You would have to replace the right hand side with something like
a_n =
{2; n <= 1
{sqrt(2/[n-1]); n > 1