Real Analysis, Lecture 12: Relationship of compact sets to closed sets

[FunkyFlashlight]

thank you :)

[mappingtheshit]

0:34:50 I like that Jing Jng

[chessnut1987]

thank you



[inf0phreak]

48:30 can any set be given a metric s.t. that it's compact? No. A compact metric space is seperable, and metric spaces are 1st-countable Hausdorff. It's a theorem that a seperable 1st-countable Hausdorff space has at most the cardinality of the continuum. Hence I^I, the set of functions from the unit interval to the unit interval, does not admit a metric under which it is compact --- though it is compact in the product topology by Tychonoff's theorem.

[mappingtheshit]

@inf0phreak Jesus Christ, your know so much, smart boy, so the question to you is why you are here ??? To show your smartness, dumbass?

[anirudh215]

During the part where he proves all nested closed intervals in R are not empty, to show that sup {a[i]} is less than all b[i], run the argument of "Every bounded set of reals has a supremum." That is, The set of all b[i] is bounded below by the a[i] and thus must have an infimum.