# Have scratchwork. Need to finish proof of sup E = inf U

1. Sep 28, 2008

### pzzldstudent

This problem was an in-class example that became assigned as additional homework. The scratchwork was done in class, but we ran out of time to write out the formal proof.

Claim: Let E be a subset of R, E is not empty and E is bounded above. Set U = {b in R: b is an upper bound of E}. Prove that sup E = inf U.

Scratchwork:
• U is not empty because E is bounded above
• If b is an upper bound of E (i.e. b is in U), then sup E ≤ b, by the definiton of supremum. Therefore sup E is a lower bound of U.
• To show it is an infimum: if ß > (what we think is the inf) and show there is an x in U such that ß > x ≥ (what we think is the infimum)
• Let ß > sup E. Need an upper bound of E (say b*) where ß > b* ≥ sup E
• Take b* = sup E

My attempt of a writeup of a proof: Let E be a subset of R, E is not empty, and E is bounded above. U is not empty because E is bounded above. If U is the set of all upper bounds, there is an element b that is an upper bound of E (i.e. b in U). Then sup E ≤ b by the definition of supremum. Therefore, sup E is a lower bound of U. Let ß > sup E. We need an upper bound of E (say b*) where ß > b* ≥ sup E. Take b* = sup E. Then by the definition of infimum, sup E = inf U.

Is that even close to a decent proof? This class is really killing me. :|
Any help at all is greatly appreciated.