Statement
If $X$ is a totally ordered set and if $\mathfrak{X}=\{X_i\subseteq X:i=1,...,n\}$ is a finite subcollection of not empty subset of $X$ with supremum then $\bigcup\mathfrak{X}$ is limited above and $\sup\{\bigcup\mathfrak{X}\}=\max\{\sup X_i\}$.
Unfortunately I can't prove the statement so I ask to do it. So could someone help me, please?
First of all we remember that every not empty finite subset of a totally ordered set has a maximum and minimum: here the proof. So we can define $\xi:=\max\{\sup X_i\}$ so that for any $x\in\bigcup\mathfrak{X}$ it follows that $x\le\xi$ and so $\bigcup\mathfrak{X}$ is limited above; but if $\alpha$ is an upper bound for $\bigcup\mathfrak{X}$ then for $i=1,...,n$ such that $\sup X_i=\xi$ it follows that $x_i\le\alpha$ for any $x_i\in X_i$ and so by the definition of $\xi$ it must be $\xi\le\alpha$ so that we conclude $\sup\bigcup\mathfrak{X}=\xi$.