Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $A$, $B$ are bounded subsets of $\Bbb R$. Prove

  1. $A\cup B$ is bounded.
  2. $\sup(A \cup B) =\sup\{\sup A, \sup B\}$.

Can anyone help with this proof?

share|improve this question
1  
(1) is very, very easy, and (2) isn’t much harder; do you have any ideas at all about how to proceed? –  Brian M. Scott Oct 20 '12 at 17:42
    
I am very positive that I got 1. But I am not sure how to start with the (2) one. –  Markus Xero Oct 20 '12 at 17:56

2 Answers 2

Without loss of generality assume that $\sup A\le\sup B$, so that $\sup\{\sup A,\sup B\}=\sup B$, and you simply want to show that $\sup(A\cup B)=\sup B$. Clearly $\sup(A\cup B)\ge\sup B$, so it suffices to show that $\sup(A\cup B)\le\sup B$.

To show that $\sup(A\cup B)\le\sup B$, just prove that $\sup B$ is an upper bound for $A\cup B$, i.e., that $x\le\sup B$ for every $x\in A\cup B$. This isn’t hard if you remember that we assumed at the start that $\sup A\le\sup B$.

share|improve this answer

for $1$ use the fact that $x\in A\cup B \Leftrightarrow x\in A$ or $x\in B$ (notice that $SupA,\space SupB$ exists since $A,\space B$ are bounded) and for $2$ use the least upper bound property. that if $SupA = M \Leftrightarrow \forall x\in A,\space \exists M\in \mathbb{R}$ such that $x\leq M$ and $\forall \epsilon>0,\space M - \epsilon \leq x$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.