# Intro to Real Analysis Proof

1. Sep 28, 2008

### pzzldstudent

I am really having a hard time in this intro to real analysis class. I feel as if I'm the only one in class who isn't getting it. I have an extremely hard time thinking abstractly and constructing my own proofs. I know I need a lot of practice. Here is the problem we have to prove:

Claim: Let A be a nonempty subset of R (all real numbers -- how do I type the symbol for real numbers?). If α = sup A is finite, show that for each ε > 0, there is an a in A such that α – ε < a ≤ α.

My attempt of a proof: Assume α = sup A is finite. Then A is bounded above because it is not empty and its supremum is finite (by the definition that if E is a nonempty subset of R (all reals), we set sup E = ∞ if E is not bounded above). [my question is where does the “ε” come from?] By definition of supremum, there is an element ß in R such that ß < α and ß is not an upper bound. In this case let ε be the ß where ε > 0. Knowing α is the supremum, ε < α, so there is an element a in A such that ε < a ≤ α or α – ε < a ≤ α.

*I also need to prove the converse of this statement which is:
"Let A be a nonempty subset of R (all real numbers) that is bounded above by α. Prove that if for every ε > 0 there is an a in A such that α – ε < a ≤ α, then α = sup A."

When proving the converse, isn't it just basically working backwards?
So I would write: Assume that for every ε > 0 there is an a in A such that α – ε < a ≤ α.
A is nonempty and bounded above by α (given). Then α = sup A is finite by the definition of supremum.

I feel really confused and lost here. I'm really afraid of this class. I need to pass it because it is only offered every 2 years.

Any help, suggestions, and guidance is greatly appreciated.
Thank you.

Last edited: Sep 28, 2008
2. Sep 28, 2008

### HallsofIvy

Staff Emeritus
Your basic idea is good but you cannot say "let $\epsilon$ be" something. You have to show that this is true no matter what $\epsilon$ is. I would have started a little differently:
Given any $\epsilon> 0$, α- $\epsilon$< α so is not an upper bound on A. Since it is not an upperbound, there exist x in A such that x> α-$\epsilon$.

3. Oct 10, 2010

### aerokat25

I have to prove this same question for my real analysis class. My is at the graduate level and I feel like a complete idiot (however, I know I am not) Help me too.