# Supremum of an union of bounded sets

Given $A$, $B$ are bounded subsets of $\Bbb R$. Prove

1. $A\cup B$ is bounded.
2. $\sup(A \cup B) =\sup\{\sup A, \sup B\}$.

Can anyone help with this proof?

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(1) is very, very easy, and (2) isn’t much harder; do you have any ideas at all about how to proceed? – Brian M. Scott Oct 20 '12 at 17:42
I am very positive that I got 1. But I am not sure how to start with the (2) one. – Markus Xero Oct 20 '12 at 17:56

Without loss of generality assume that $\sup A\le\sup B$, so that $\sup\{\sup A,\sup B\}=\sup B$, and you simply want to show that $\sup(A\cup B)=\sup B$. Clearly $\sup(A\cup B)\ge\sup B$, so it suffices to show that $\sup(A\cup B)\le\sup B$.
To show that $\sup(A\cup B)\le\sup B$, just prove that $\sup B$ is an upper bound for $A\cup B$, i.e., that $x\le\sup B$ for every $x\in A\cup B$. This isn’t hard if you remember that we assumed at the start that $\sup A\le\sup B$.
for $1$ use the fact that $x\in A\cup B \Leftrightarrow x\in A$ or $x\in B$ (notice that $SupA,\space SupB$ exists since $A,\space B$ are bounded) and for $2$ use the least upper bound property. that if $SupA = M \Leftrightarrow \forall x\in A,\space \exists M\in \mathbb{R}$ such that $x\leq M$ and $\forall \epsilon>0,\space M - \epsilon \leq x$