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
Proof of the Sequence Limit Law for Difference of Sequences: https://youtu.be/ZlAT0_W4yP0
Proving All the Sequence Limit Laws: https://youtu.be/dWLhfYUb3wc
Proof of the Sequence Squeeze Theorem: https://youtu.be/hYY9piVkB6g
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