# 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.