TITLE: Supremums and subsets
QUESTION [2 upvotes]: Let $A \subseteq B$, where $A$ and $B$ are non-empty sets and $B$ is bounded above.
Show that $\sup(A)$ exists and $\sup(A)\le \sup(B)$ (Problem given as written)
For clarity, $\sup(x)$ is the supremum of x
I'm just asking if I did this correctly, I feel like I'm missing something here
Proof sup(A) exists:
Let $Y = \max(B)$. Then, $Y \ge a, \forall a\in A$.
Therefore, A is bounded above because $\exists b\in B$ such that $b \ge a,\forall a\in A$.
By completeness axiom, if A is bounded above, then $\exists \sup(a)$
Proof $\sup(A)\le\sup(B)$: Since $\sup(B)$ is an upper bound of $B$ and $A\subseteq B$, $\sup(B)$ is an upper bound of $A$ such that $\sup(A) = \sup(B)$ or $\sup(A) < \sup(B)\implies \sup(A)\le \sup(B)$.

REPLY [1 votes]: The existence part is very close, but you can't let $Y$ be the maximum of $B$, because $B$ might not have a maximum!  But it's bounded above, and if it is a subset of $\Bbb R$, then the completeness of $\Bbb R$ implies it has a least upper bound, i.e., $\sup{B}$ exists.  If you let $Y = \sup{B}$ instead, then your argument for existence works perfectly.  You show $Y$ is a bound for $A$, as you did, and then by completeness of $\Bbb R$, then $A$ is bounded from above and so has a least upper bound, i.e., $\sup{A}$ exists.
Your argument for the second part is also fine.  Since $\sup{B}$ is an upper bound for $A$, and $\sup{A}$ is the least upper bound, it follows that $\sup{A} \leq \sup{B}$.  Good job.

Just to add a comment: I wanted to give you an example of a subset of $\Bbb R$ that has a supremum but not a maximum.  Take $A = (0, \pi) \cap \Bbb Q$.  Clearly, this is bounded, so it has a least upper bound (which is actually $\pi$).  But can you find a largest element in this set?