# let $A_1, A_2, A_3, \dots$ be a collection of nonempty sets, each of which is bounded above. $(a)$ Find a formula for $\sup(A_1\cup A_2)$.

Let $$A_1, A_2, A_3,\dots$$ be a collection of nonempty sets, each of which is bounded above.

$$(a)$$ Find a formula for $$\sup(A_1 \cup A_2)$$. Extend this to supremum of a collection of $$n$$ sets $$A_1, A_2, \dots, A_k$$.

For $$(a)$$ I want to say that it's just the largest of the supremums, but I'm not sure how to show or prove that.

$$(b)$$ Consider the supremum of an infinite number of sets. Does the formula in $$(a)$$ extend to the infinite case?

For $$(b)$$ is it possible to have a supremum of an infinite number of sets as long as they're all bounded above?

• Welcome to Mathematics Stack Exchange.
– user249332
Sep 6, 2015 at 19:01
• Are these subsets of the real line $\Bbb R$? Sep 6, 2015 at 19:03
• Yes they are Berci Sep 6, 2015 at 19:05
• I think $\sup(A_1 \cup A_2)=\sup(\{\sup(A_1),\sup(A_2)\})$. Sep 6, 2015 at 19:07
• For part (a) : math.stackexchange.com/questions/921975/… Oct 6, 2017 at 12:16

a) Yes, we have $\ \sup(A_1\cup A_2)=\max(\sup A_1,\,\sup A_2)$.
To prove this, just use the definition of $\sup$ (e.g. for a subset $U$ and an element $v$ we have $U\le v\iff \sup U\le v\$ where $U\le v$ wants to mean $u\le v$ for all $u\in U$.)
b) Well, the supremum can also be $+\infty$, and yes, it is possible to achieve, for a simplest example take $A_n:=\{n\}$. Each of these sets is of course bounded, but their union is not.
Can you find the formula for $\sup(A_1\cup A_2\cup\dots)$?
• Yes, there is formula (if we also allow $+\infty$, but why not?). And that formula is basically on this page. Sep 7, 2015 at 20:10