Question

Let A and B be non-empty, bounded subsets of R. (a) Why does sup(AUB) exist? (b) Prove that sup(AUB) = max{sup A, sup B}.

Answer 1

Answer with Step-by-step explanation:

Let A and B be non- empty bounded subset of R

a.We have to find why $sup(A\cup B)$exist

If A and B are bounded set

Then there exist  constant  such that

$a\leq A\leq b$ and $c\leq B\leq d$

Then , sup of A =b and sup of B=d

When  a set is bounded then all elements lie in the set are lie between the constants s and t.

All elements are less than or equal to t then t is supremum of set.

Because both set are bounded and sup of both set A and B are exist.All elements A union B are less than or equal to sup A or sup B.

$sup(A\cup B)=max(sup A, sup B)$

Then, $sup (A\cup B)$ exist.

b.We have to prove that

$sup (A\cup B)=max(sup A,sup B)$

Suppose ,A =(1,2) and B=(2,3)

Sup A=2 , sup B=3

$(A\cup B)=(1,2)\cup (2,3)$

Upper bound of $A\cup B)=3$

Hence, $Sup (A\cup B)=3$

If A=(4,5),B=(2,3)

Sup A=5,Sup B=3

$A\cup B=(4,5)\cup (2,3)$

$Sup(A\cup B)=5$

Hence, $Sup(A\cup B)=5$

Hence, we can say that $sup(A\cup B)=max(sup A,sup B)$.