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$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.

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  • $\begingroup$ To clarify: I assume that the sets you are talking about are sets of real numbers ? $\endgroup$ – Tom Collinge Aug 29 '15 at 14:24
  • $\begingroup$ yes, correct! Real numbers $\endgroup$ – Laura Aug 29 '15 at 14:26
  • $\begingroup$ Consider that $\bigcup^{n+1}_{k=1}A_k = (\bigcup^{n}_{k=1}A_k) \bigcup A_{k+1}$ $\endgroup$ – Tom Collinge Aug 29 '15 at 14:28
  • $\begingroup$ @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}$} ? $\endgroup$ – Laura Aug 29 '15 at 14:41
  • $\begingroup$ 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. $\endgroup$ – Tom Collinge Aug 29 '15 at 14:48
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$\sup(A \cup B \cup C) = \sup( (A \cup B) \cup C ) = \max( \sup(A \cup B) , \sup(C) )$ [by the 2-set case]

$\ = \max( \max( \sup(A) , \sup(B) ) , \sup(C) )$ [again] $ = \max( \sup(A) , \sup(B) , \sup(C) )$.

I'm sure you can see how to extend this to arbitrary finite number of sets!

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