# $\sup A \le \beta$ proof verification

If $A \subset B, a_0 \in A$ and $\beta$ is an upper bound of B then $\sup A \le \beta.$

$\textbf{Proof:}$

Since $a_0 \in A,$ then $A \ne \emptyset,$ thus $\sup A$ exists. Since $A \subset B,$ then $\sup B$ is an upper bound of $A$. By definition, $\sup A$ is the least upper bound of $A.$ This tells us that $\sup A \le \sup B.$ Also, since $\beta$ is an upper bound of $B,$ then $\sup B \le \beta.$ This implies $\sup A \le \sup B \le \beta,$ hence $\sup A \le \beta.$

Please tell me if this is a correct way in proving such a statement. Any constructive criticism would be much appreciated. Thank you very much.

• yes you are correct Feb 21, 2015 at 6:44
• Awesome! thank you!!! Feb 21, 2015 at 6:45
• Do you allow $+\infty$ as a value for $\sup A$, or does the definition require this to be a real number? If the latter, then the first sentence in your proof is unjustified. If the former, you should probably clarify to avoid misunderstandings. Feb 21, 2015 at 7:03
• @ Andres Caicedo I believe in Rudin, the $sup \; A$ is allowed to be $+\infty$ Feb 21, 2015 at 7:06
• You can use $\sup A$, which renders as $\sup A$. It looks a bit better and, most importantly, it has correct spacing. (It is not all lumped together as in $sup A$ $sup A$.) Feb 21, 2015 at 8:32

I agree. It seems fine. But I'm wondering if you can't just say that, if $\beta$ is an upper bound for $B$ and $A\subseteq B$, then $\beta$ is an upper bound for $A$ and hence $\sup A\leq\beta$. Maybe we have to be extra careful because of the versions of the definitions we have in mind? Saying that $A\subseteq B\Rightarrow \sup A \leq \sup B$ seems like a stronger thing to use in your reasoning than just $A\subseteq B\Rightarrow$ "an upper bound for $B$ is an upper bound for $A$" and "$\beta$ is an upper bound for $A$" $\Rightarrow \sup A \leq \beta$. Think about it and let me know what you think. :)