 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Let $M=\max\{\sup A,\sup B\}$. Every element of the <union> $A\cup B$ is at most $M$, so $M$ is an <upper bound>. Conversely, every <upper bound> of $A\cup B$ is an <upper bound> of each of $A$ and $B$, hence is at least both <suprema> and therefore at least $M$. Consequently
$$
\boxed{\sup(A\cup B)=\max\{\sup A,\sup B\}.}
$$

 Back to article page