Am I correctly using the properties of the supremum in this proof?

[jdinatale]

Hi, I was wondering if I correctly applied the properties of the supremum of a set to solve the following proof. I feel like I "cheated" in the sense that I said, "Let s = Sup(B) - epsilon.

Homework Statement

If $\sup A < \sup B$, then show that there exists an element $b \in B$ that is an upper bound for $\sup A$.

Homework Equations

None

The Attempt at a Solution

 

[Tedjn]

You are essentially correct but here are some thoughts. This element s you claim is any arbitrary upper bound for A. Being any arbitrary upper bound for A there is no guarantee that s is less than sup B (hence epsilon is positive). You would need to argue that such an s exists between sup A and sup B. This isn't hard. But take this even further: recall sup A itself is already an upper bound for A, so instead of using this extra s, your proof works fine using just sup A (which we don't need to argue is less than sup B because it is given to us).

Your intuition that you cheated is so probably because there are a few small steps in reasoning jumped over by invoking lemma 1.3.7. If you think through lemma 1.3.7 as well, I believe you will feel more confident in your argument. (The crux of the lemma, of course, is that sup B is the least upper bound for B, hence there must be elements in B arbitrarily close to sup B.)

 

[jdinatale]

You are essentially correct but here are some thoughts. This element s you claim is any arbitrary upper bound for A. Being any arbitrary upper bound for A there is no guarantee that s is less than sup B (hence epsilon is positive). You would need to argue that such an s exists between sup A and sup B. This isn't hard. But take this even further: recall sup A itself is already an upper bound for A, so instead of using this extra s, your proof works fine using just sup A (which we don't need to argue is less than sup B because it is given to us).

Your intuition that you cheated is so probably because there are a few small steps in reasoning jumped over by invoking lemma 1.3.7. If you think through lemma 1.3.7 as well, I believe you will feel more confident in your argument. (The crux of the lemma, of course, is that sup B is the least upper bound for B, hence there must be elements in B arbitrarily close to sup B.)

Thank you for taking the time to help me. I'm not sure what you mean by disregard s. I reworked my proof and found I needed to use an s. What are your thoughts on the new proof?

 

[Tedjn]

Yes, this works. Here is what I meant by not using s, however. See if this also makes sense as a simplification:

We know sup A < sup B. Let ε = sup B - sup A, which is positive. By Lemma 1.3.7, there exists a b ∈ B such that b > sup B - ε = sup A. Thus, b is an upper bound for A.