Lecture 3: Math. Analysis - Bounded sets

[@kanikabagree1084]

Thankyou so much for the amazing lecture!

[@christinedunnett6690]

If you listen to this in double speed, it sounds like Jimmy Carr.

[@postbodzapism]

Can I say that because A is a subset of B iff (x e A => x e B) where e means "is an element of", and (x e A => x e B) is equivalent to (x !e B => x !e A) where !e means "is not an element of", and that if x !e some set X, then x !e empty set because no x is an element of the empty set, so (x !e X => x !e empty set) holds for any set X, so the empty set is a subset of any set X?

[@ericcooke2661]

Yes, it also implies that if there exists x !E B, then B must be a subset of another set of U.

[@rikbrutsaert5053]

This like pulling teeth. Need more advanced examples.

[@uniofnottingham]

***** Students are often confused by the term “bounded”, so lots of easy examples are needed. My comments from 4:00 onwards are relevant here. See also my Blog post wp.me/posHB-v - Dr Joel Feinstein

[@rikbrutsaert5053]

***** Prof. Feinstein, I see what you are saying! I am currently a master's student at Cornell in stats and they want me to do well in honors analysis to get into the doctorate program so I can write proofs. The examples we are getting in my course are pretty advanced (the honors version of the class is for the "go-getters"), and I've been out of school for a number of years so I lack the foundational base. It's challenging... Anyway, keep up the good work, I appreciate the video!