Proofs of Supremum Problem in $\mathbb{R}$

[alexmahone]

(a) Let $S$ be a bounded non-empty subset of $\mathbb{R}$, and $\overline{m}=\sup S$. Prove there is a sequence $\{a_n\}$ such that $a_n\in S$ for all $n$, and $a_n\to\overline{m}$. (You must show how to construct the sequence $a_n$.)

(b) Let $A$ and $B$ be bounded non-empty subsets of $\mathbb{R}$. Prove the equality $\sup (A+B)=\sup A+\sup B$. (Use part (a).)

 

[Opalg]

(a) Let $S$ be a bounded non-empty subset of $\mathbb{R}$, and $\overline{m}=\sup S$. Prove there is a sequence $\{a_n\}$ such that $a_n\in S$ for all $n$, and $a_n\to\overline{m}$. (You must show how to construct the sequence $a_n$.)

(b) Let $A$ and $B$ be bounded non-empty subsets of $\mathbb{R}$. Prove the equality $\sup (A+B)=\sup A+\sup B$. (Use part (a).)

The definition of sup tells you that, for each $n\geq1$, $\overline{m}-\frac1n$ is not an upper bound for $S$. So there exists an element $a_n\in S$ with $a_n>\overline{m}-\frac1n$. That gives you your sequence $\{a_n\}$.

 

[alexmahone]

Thanks. How about part (b)?

 

[Opalg]

Thanks. How about part (b)?

Use the hint. If $a_n\to\sup A$ and $b_n\to\sup B$, what can you say about the sequence $\{a_n+b_n\}$?

 

[alexmahone]

Use the hint. If $a_n\to\sup A$ and $b_n\to\sup B$, what can you say about the sequence $\{a_n+b_n\}$?

$\{a_n+b_n\}\to\sup A+\sup B$

How do I proceed?

 

[Opalg]

$\{a_n+b_n\}\to\sup A+\sup B$

How do I proceed?

Since $a_n+b_n\in A+B$, that shows that $\sup(A+B)\geq\sup A+\sup B$. What about the reverse inequality?

 

[alexmahone]

Since $a_n+b_n\in A+B$, that shows that $\sup(A+B)\geq\sup A+\sup B$. What about the reverse inequality?

I know how to prove the reverse inequality. But I'm not sure I understand how you got $\sup(A+B)\geq\sup A+\sup B$?

 

[Opalg]

Since $a_n+b_n\in A+B$, that shows that $\sup(A+B)\geq\sup A+\sup B$. What about the reverse inequality?

I know how to prove the reverse inequality. But I'm not sure I understand how you got $\sup(A+B)\geq\sup A+\sup B$?

From $a_n+b_n\in A+B$ it follows that $a_n+b_n\leq\sup(A+B)$. Now let $n\to\infty$ to get $\sup A + \sup B \leq \sup(A+B)$.

 

[alexmahone]

From $a_n+b_n\in A+B$ it follows that $a_n+b_n\leq\sup(A+B)$.

While I agree with this inequality for all $n\in\mathbb{N}$, I don't understand how we may let $n$ tend to $\infty$. After all, the values $\sup A$ and $\sup B$ may never be attained by $a_n$ and $b_n$ for any $n\in\mathbb{N}$.

Sorry if I'm being slow but I'm quite new to analysis.

 

[Opalg]

You need to use the fact the weak inequalities are preserved by limits. If $x_n \leq L$ for all $n$, then $\lim_{n\to\infty}x_n \leq L$.

 

[alexmahone]

You need to use the fact the weak inequalities are preserved by limits. If $x_n \leq L$ for all $n$, then $\lim_{n\to\infty}x_n \leq L$.

Ah -- the Limit location theorem. It tells me that $\lim (a_n+b_n)\le\sup(A+B)$, which is the same as $\sup A+\sup B\le\sup(A+B)$.

Thanks a ton, Opalg! No wonder you were voted Best Analyst on MHF.