Ep. 4 Limit Points, LimSup, and Completeness of R^n: Navigating Metric Space Topology

We define the concept of limsup of a bounded real sequence and prove that it is the maximum limit point. We also explore the limit of a function between metric spaces. Finally, we wrap up by introducing Cauchy sequences and then showing that R (and more generally R^n) is a complete metric space.

* Errata: There is an error in the argument that no limit point is larger than limsup (starting 15:18). Note that a_{n_1}, a_{n_2}, ... in the proof does not need to be monotonically increasing, so it's incorrect that all b_{n_k} are at least a_{n_1}, as incorrectly stated in 16:52. The fix is simple: Since a_{n_1}, a_{n_2}, ... is converging to b' and b' is strictly larger than b, we can pick some real number c strictly larger than b such that all terms of this subsequence (after possibly discarding initial terms) lie above c. Now, using this value c rather than a_{n_1}, the exact same argument as in the video goes through.

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[LetsSolveMathProblems]

* Errata: There is an error in the argument that no limit point is larger than limsup (starting 15:18). Note that a_{n_1}, a_{n_2}, ... in the proof does *not* need to be monotonically increasing, so it's incorrect that all b_{n_k} are at least a_{n_1}, as incorrectly stated in 16:52. The fix is simple: Since a_{n_1}, a_{n_2}, ... is converging to b' and b' > b, we can pick some real number c > b such that all terms of this subsequence (after possibly discarding initial terms) lie above c. Now, using this value c rather than a_{n_1}, the exact same argument as in the video goes through.

[andrewdsotomayor]

Always been told I need to learn my limits, here’s my chance!

[zmaj]

Problem 4.
Choose eps > 0. There exists some N such that, for all k>=N, | sum_{n=1 to k} (a_n) - a | < eps/2. Then, for any a_i, where i>=N+1, | sum_{n=1 to i-1} (a_n) - a | < eps/2, and | a - sum_{n=1 to i} (a_n) | < eps/2. Using the triangle inequality, we find that |a_i - 0| = | (sum_{n=1 to i-1} (a_n) - a) + (a - sum_{n=1 to i} (a_n)) | <= | sum_{n=1 to i-1} (a_n) - a | + | a - sum_{n=1 to i} (a_n) | < eps. From this, we gather that the sequence {a_n} converges to 0.