# Help! Intro to Real Analysis problem

• Nov 18th 2008, 10:43 AM
Robitar
Help! Intro to Real Analysis problem
I'm having problems with some of the proofs of the supremum and infimum properties in my text. here are two that I'm stuck on:

a) Let S be a non-empty subset of R that is bounded below. Prove that
inf S = -sup{-s: s in S}

b) Let A and B be bounded non-empty subsets of R, and let
A+B := {a+b: a in A, b in B}.
Prove that sup(A+B) = supA + supB and inf(A+B) = infA + infB

Any comments and/or help would be much appreciated, it's getting near the end of the semester, and I'm really struggling with these concepts!

-Robitar
• Nov 18th 2008, 12:09 PM
HallsofIvy
Quote:

Originally Posted by Robitar
I'm having problems with some of the proofs of the supremum and infimum properties in my text. here are two that I'm stuck on:

a) Let S be a non-empty subset of R that is bounded below. Prove that
inf S = -sup{-s: s in S}

"inf S" is defined as the largest of all lower bounds on S. Let a= inf S. Then a is a lower bound on S so if x is any member of S, then a< x. Can you use that to show that a is an upper bound on {-s: s in S}? Try multiplying by -1. If y is any other lower bound on S, then a< y. Prove that -y is also an upper bound on {-s: s in S} and -a> -y.

Quote:

b) Let A and B be bounded non-empty subsets of R, and let
A+B := {a+b: a in A, b in B}.
Prove that sup(A+B) = supA + supB and inf(A+B) = infA + infB
If x is in A+ B, then there exist a in A and b in B such that x= a+ b.
Show that a< sup(A) and b< sup(B) so a+b< sup(A)+ sup(B) so sup(A)+ sup(B) is an upperbound on A+ B. What does that prove about sup(A+ B), the least upperbound on A+ B?
Any comments and/or help would be much appreciated, it's getting near the end of the semester, and I'm really struggling with these concepts!

-Robitar[/QUOTE]
• Nov 18th 2008, 01:54 PM
Mathstud28
Quote:

Originally Posted by Robitar

a) Let S be a non-empty subset of R that is bounded below. Prove that
inf S = -sup{-s: s in S}

Define $A=\left(a_1,a_2,\cdots,a_n\right)$ and $-A=\left(-a_1,-a_2,\cdots,-a_n\right)=B$. So your question is equivalent to showing that $\inf\left(A\right)=-\sup\left(B\right)$.

1. Since $A\subset\mathbb{R}$, it is nonempty, and bounded we have that $\inf\left(A\right)$ exists since $\mathbb{R}$ has the greatest lower bound property. Furthermore since $\mathbb{R}$ is an ordered field we have that $a_1\leqslant{a_2}\leqslant\cdots\leqslant{a_n}$. So we know that $\forall{\gamma}\in{A}~a_1\leq{\gamma}$, where it follows that $a_1$ is a lower bound of $A$. Now it is pretty obvious that if $a_1<\gamma$ then $\gamma$ is not a lower bound of $A$. Therefore we conclude that $a_1$ is a greatest lower bound of $A$, or $\inf\left(A\right)=a_1$.

2. Now consider once again $a_1\leqslant{a_2}\leqslant\cdots\leqslant{a_n}$, now by the ordering of $\mathbb{R}$ we would have that $-a_1\geqslant{-a_2}\geqslant\cdots\geqslant{-a_n}$, so we can see that since $B=\left(-a_1,-a_2,\cdots,-a_n\right)$ that $\forall{\gamma}\in{B}~\gamma\leqslant{-a_1}$, so $B$ is bounded above by $-a_1$. Now noting that $B=-A$ gives us two things, $B\subset\mathbb{R}$ and $B\ne\varnothing$, so once again because $\mathbb{R}$ has the leas upper bound property we can conclude that $\sup\left(B\right)$ exists. Now as noted before $-a_1$ is an upper bound of $B$, and it can be seen that if $\gamma<-a_1$ then $\gamma$ is not an upper bound of $B$. So we can see that $-a_1$ is a least upper bound of $B$ or in other words $\sup\left(B\right)=-a_1$

3. So conclude that since $\inf\left(A\right)=a_1$ and $\sup\left(B\right)=-a_1$ that $\inf\left(A\right)=-\sup\left(B\right)~\blacksquare$