MATH 242: Analysis 1 (McGill Final)

Currently Playing: Bounded Set Contains Sequence Converging to its Supremum | Real Analysis

Support the production of this course by joining Wrath of Math to access all my real analysis videos plus the lecture notes at the premium tier! https://www.youtube.com/channel/UCyEKvaxi8mt9FMc62MHcliw/join 🛍 Get the coolest math clothes in the world! https://mathshion.com/ Real Analysis course: https://www.youtube.com/playlist?list=PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli Real Analysis exercises: https://www.youtube.com/playlist?list=PLztBpqftvzxXAN05Gm3iNmpz9SkVfLNqC Get the textbook! https://amzn.to/3CMdgjI Any bounded subset of the real numbers contains a sequence converging to its supremum. This is a nice connection we can make between supremums and sequence limits. Speaking precisely, we know that a subset S of the real numbers with an upper bound has a supremum by the completeness axiom. Call this supremum s. Then, there exists a sequence a_n, where each term of the sequence is in S, and a_n converges to s. We prove this result using the epsilon definition of supremum, the sequence squeeze theorem, and some other basic limit laws. Epsilon Definition of Supremum: https://studio.youtube.com/video/lCpoEWJRYNk/edit Constant Sequence Converges to its Constant Value: https://youtu.be/O-quXJi6-ws Follow Wrath of Math on... ● Instagram: https://www.instagram.com/wrathofmathedu ● Facebook: https://www.facebook.com/WrathofMath ● Twitter: https://twitter.com/wrathofmathedu