Thread: Check this proof on supremums?

1. Check this proof on supremums?

This is from Fitzpatrick's Advanced Calculus. I am not very experienced with proofs so could someone tell me if this is a correct way of doing it? Thanks!

Claim:
If $S$ is a set of real numbers that is bounded above and $B$ is a nonempty subset of $S$, then $\text{sup }B\leq \text{sup }S$.

Proof: Let $S$ be a set of real numbers that is bounded above and $B$ a nonempty subset of $S$. Take an arbitrary element $x$ in $B$. Since $B\subseteq S$, $x$ is also in $S$. Since $S$ is bounded above, by the completeness axiom we know there is a number $c=\text{sup }S$ such that $x\leq c$, and there is no number less than $c$ that is an upper bound for $S$. Since $x$ is in $B$, $c$ is also an upper bound for $B$. So we must have $\text{sup }B\leq c$, or $\text{sup }B\leq \text{sup }S$.

2. Re: Check this proof on supremums?

Hey Ragnarok.

Does your professor require you to use the full definition of the supremum? (i.e. the supremum about being the lowest greater bound?)

3. Re: Check this proof on supremums?

Chiro is correct to refer you to your professor - there are probably certain things that you need to make sure you include.

I would write a lot less to prove this claim: Let $x\in{B}$. Then since B is a subset of S, $x\in{S}$, so $x\le\sup{S}$. Since x was an arbitrary element of B, $\sup{B}\le\sup{S}$.

But again, your professor might want more.

- Hollywood

4. Re: Check this proof on supremums?

Thanks guys! I'm actually self-studying, so I don't have a professor to refer to. I'm just guessing on the definitions and form. Your shortened version helps me a lot.