1. ## Supremum

Suppose A is a subset of R bounded above and suppose we are given that c "is an element of" R.

Define the sets c + A and cA by c + A = {c + a : a "is an element of" A} and cA = {ca : a "is an element of" A}.

1.) Show that sup(c + A) = c + sup(A)

2.) If c >= 0, show that sup(cA) = c*sup(A)

3.) Postulate a similar statement for sup(cA) for the case where c < 0.

2. Originally Posted by seerTneerGevoLI
Suppose A is a subset of R bounded above and suppose we are given that c "is an element of" R.

Define the sets c + A and cA by c + A = {c + a : a "is an element of" A} and cA = {ca : a "is an element of" A}.

1.) Show that sup(c + A) = c + sup(A)

2.) If c >= 0, show that sup(cA) = c*sup(A)

3.) Postulate a similar statement for sup(cA) for the case where c < 0.
Here is a stronger theorem.

Let A and B be non-empty subsets of R. Define A+B to be {a+b | a in A and b in B}. Show that if A and B have an upper bound then A+B has an upper bound with the property that:
sup(A+B) = sup(A) + sup(B)

Can you show that?

3. Not really.

There being a "least upper bound" and a "greatest lower bound" is a little confusing.

For this, a set A "is a subset of" R is bounded above if there exists a number b belonging to R such that a <= b for all a belonging to A.

And, I know that a real number s is the least upper bound for a set A "is a subset of" R if:

1.) s is *AN* upperbound for A <-- how there can be more than 1 upperbound is still beyond me

2.) b is any upper bound for A, then s <= b

So for this problem, by defining c + A and cA as above, if a is an element of A in both those situations, it's not dependent on c. So this is how I see why sup(c + A) = c + sup(A). But just because it's independent does not make it independent of the least upper bound?

Now for #2, if c >= 0, then it would make sense that sup(cA) = c*sup(A) for the same reason as above.

But I'm not sure what they want in formarly showing this.

And lastly, if c < 0, wouldn't this just be sup(cA) = c*inf(A), as it's no longer a least upper bound, but a greatest lower bound as a result of the negative?

4. Theorem: Let $A \mbox{ and }B$ be non-empty subsets of $\mathbb{R}$. Define $A+B = \{ a + b | a\in A \mbox{ and }b\in B\}$. If $A,B$ have an upper bound then $A+B$ has an upper bound. And furthermore, $\sup (A+B) = \sup (A)+\sup (B)$.

(If you are new to this proof stuff do the following: Read each sentence and try to explain to yourself each detail that I state.)

Proof:
(Part I)Notice that $a\leq \sup(A)$ for all $a\in A$. And $b\leq \sup(B)$ for all $b\in B$. That means $a+b\leq \sup(A) + \sup(B)$. Hence $A+B$ has an upper bound $\sup(A)+\sup(B)$. Thus by completeness $\sup(A+B)$ exists and furthermore, $\boxed{ \sup(A+B)\leq \sup(A)+\sup(B) }$.
(Part II)Notice that $a+b\leq \sup(A+B)$ for all $a\in A, \ b\in B$. Thus, $a\leq \sup(A+B)-b$ for any $b$ and for all $a$. Thus, $\sup(A)\leq \sup(A+B) - b$ for any $b$. Thus, $\sup(A+B) - \sup(A)\geq b$ for any $b$. Thus, $\sup(A+B)-\sup(A)\geq \sup(B)$. Thus, $\boxed{ \sup(A+B)\geq \sup(A)+\sup(B) }$.
Q.E.D.

Do you understand this proof?

Now you can easily prove that $\sup \{k+a|a\in A\} = k +\sup (A)$. Because we can think of $k$ as a singleton set $B=\{ k \}$ and $A$ as itself. Then $\sup (A+B) = \sup(A)+\sup(B) = \sup(A)+k$.

5. Originally Posted by seerTneerGevoLI
2.) If c >= 0, show that sup(cA) = c*sup(A)
First I just want to know which Real Analysis book are you using?

I start you off. Let $A$ be a non-empty bounded subset of $\mathbb{R}$. Then, $a\leq \sup(A)$ for all $a\in A$. Thus, $ca\leq c\sup(A)$ since $c\geq 0$ (so the inequality is not flipped). Thus, $c\sup(A)$ is an upper bound for $cA$. Thus, $\boxed{ \sup(cA)\leq c\sup(A) }$. Now (second part) $ca\leq \sup(cA)$ for all $a\in A$. Thus, $a\leq c^{-1}\sup(cA)$ for all $a\in A$. Thus, $\sup(A)\leq c^{-1}\sup(cA)$. Thus, $c\sup(A)\leq \sup(cA)$. Thus, $\boxed{ \sup(cA)\geq c\sup(A) }$. Q.E.D.

Warning: There is one really really important detail that I purposely left out for you to find. It is so important in mathematics. You never had to worry about it in your easier classes. But since now you are studing advanced mathematics you always need to consider it. Can you find it? Because without it what I wrote it worthless. Find that detail and fix my proof.

6. Thanks TPH. That did help a lot. I have one question though. What difference does it make if c is >=0 or if c < 0 as the following ask:

2.) If c >= 0, show that sup(cA) = c*sup(A)

3.) Postulate a similar statement for sup(cA) for the case where c < 0.

Do you treat c again as being B = {c}. So,

sup(BA) = sup(B)*sup(A), as you did with the addition.

Then, if c is negative, would this negate the supremum to the infinum? That is, instead of a least upper bound, there will instead be a greatest lower bound?

EDIT: Just saw your new post; I'll read it over. Thanks again.

7. As far as my Real Analysis book, I am using: "Understanding Analysis" by Stephen Abbott.

8. Originally Posted by ThePerfectHacker

Warning: There is one really really important detail that I purposely left out for you to find. It is so important in mathematics. You never had to worry about it in your easier classes. But since now you are studing advanced mathematics you always need to consider it. Can you find it? Because without it what I wrote it worthless. Find that detail and fix my proof.
Hmm. Looks like a sound proof to me. Does it have to do with extistence and uniqueness? It's the only thing that I can think of that's quite important for virtually all problems.

9. Originally Posted by seerTneerGevoLI
Hmm. Looks like a sound proof to me. Does it have to do with extistence and uniqueness? It's the only thing that I can think of that's quite important for virtually all problems.
Let this be your first and last time you make this mistake. It leads to mathematical death! I used division by zero. How? Look at what I did. I brought $c$ other to the other side $c^{-1}$. Now $c^{-1}$ does not exist when $c=0$. So I really should have considered two cases. Case 1, is $c>0$ and you can apply my argument. Case 2, is $c=0$ and you need to apply a different argument. How will it go when $c=0$?

10. Originally Posted by ThePerfectHacker
Let this be your first and last time you make this mistake. It leads to mathematical death! I used division by zero. How? Look at what I did. I brought $c$ other to the other side $c^{-1}$. Now $c^{-1}$ does not exist when $c=0$. So I really should have considered two cases. Case 1, is $c>0$ and you can apply my argument. Case 2, is $c=0$ and you need to apply a different argument. How will it go when $c=0$?
I'm quite surprised I didn't catch that, especially after taking a course in Lin. Alg. and having to always check that.

I was working through the c < 0 case.

Does this work:

Let A be a non-empty subset of R.
Then, a <= sup(A) for all a in A.
Thus, (-1)*ca > (-1)c*sup(A) such that c > 0. We want the (-1) to make c < 0.

Then, (-1)c*sup(A) is the greatest lower bound for (-1)*ca
Implies inf(cA) < c*inf(A)

**(I'm not sure if by negating the supremum you do get the infimum - I'm assuming not as the question asks to prove the statement sup(cA) for the case where c < 0.

Let me know if this is the right track. As far as when c = 0 for your one proof, wouldn't it be obvious that sup(0) = 0?

11. Originally Posted by seerTneerGevoLI
3.) Postulate a similar statement for sup(cA) for the case where c < 0.
Theorem: $\sup(-A) = - \inf (A)$.

Try to prove this. Once you have had done:

Theorem: If $c<0$ then $\sup(cA)=c\inf(A)$.

Proof: We can write $c=(-1)k$ where $k>0$. We know that $\sup(kA) = k\sup(A)$ by previous excercise. By above theorem $\sup(cA) =\sup(-kA) = -\inf (kA) = -k\inf(A)=c\inf(A)$. Q.E.D.

Now I used the result $\inf(kA) = k\inf(A)$ for $k>0$. Which should be easy to prove by imitating the proof of $\sup(kA)=k\sup(A)$.