# Proving a formula about the supremum of a finite family of sets by induction

$A_1, A_2, A_3\dots$ are a collection of nonempty sets, each bounded above.

I'm asked to find a formula for $\sup(A_1\cup A_2)$ and then to extend this to $\bigcup^{n}_{k=1}A_k$.

For the supremum of $A_1\cup A_2$ I have shown that it is $\max\{a_1, a_2\}$ when $a_1$ and $a_2$ are the suprema of $A_1$ and $A_2$ which we know they must have because of the completeness axiom.

I'm having trouble extending it to $\bigcup^{n}_{k=1} A_k$. I am told to do so by induction. I proved it for the case $n=2$ in the first step, and I think my induction hypothesis is to suppose that it is true for $\bigcup^{n}_{k=1}A_k$ and then use this to show it is also true for $\bigcup^{n+1}_{k=1}A_k$. This is where I'm stuck.

• To clarify: I assume that the sets you are talking about are sets of real numbers ? – Tom Collinge Aug 29 '15 at 14:24
• yes, correct! Real numbers – Laura Aug 29 '15 at 14:26
• Consider that $\bigcup^{n+1}_{k=1}A_k = (\bigcup^{n}_{k=1}A_k) \bigcup A_{k+1}$ – Tom Collinge Aug 29 '15 at 14:28
• @TomCollinge Can I from there say that sup($\bigcup^{n+1}_{k=1}$$A_k)=sup(\bigcup^{n}_{k=1}$$A_k$)$\cup$sup{$A_{k+1}$} ? – Laura Aug 29 '15 at 14:41
• Yes (actually max uf the sup's, not their union): $\bigcup^{n+1}_{k=1}A_k$ is a set, so $(\bigcup^{n}_{k=1}A_k) \bigcup A_{k+1}$ is just the union of two sets, and you've already proved that result. – Tom Collinge Aug 29 '15 at 14:48

$\sup(A \cup B \cup C) = \sup( (A \cup B) \cup C ) = \max( \sup(A \cup B) , \sup(C) )$ [by the 2-set case]
$\ = \max( \max( \sup(A) , \sup(B) ) , \sup(C) )$ [again] $= \max( \sup(A) , \sup(B) , \sup(C) )$.