# Math Help - Help! Intro to Real Analysis problem

1. ## 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

2. 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.

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]

3. 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 $\inf\left(A\right)=-\sup\left(B\right)~\blacksquare$