# Real Analysis Least Upper Bound Question

utstatistics

## Homework Statement

If S1, S2 are nonempty subsets of ℝ that are bounded from above, prove that

l.u.b. {x+y : x \in S1, y \in S2 } = l.u.b. S1 + l.u.b. S2

## Homework Equations

Least Upper Bound Property

## The Attempt at a Solution

Using the least upper bound property, let us suppose that a is an upper bound for S1 and b is an upper bound for S2, as they are both a set of real numbers. Then there exists an x in S1 s.t. x≤a and there exists a y in S2 s.t. y≤b. By adding them together, x+y≤a+b where a and b are the upper bounds for their respected set of real numbers. Thus, it is equal to l.u.b. S1 + l.u.b. S2

I think this is how it goes. I'm sorry if my proof is horrible; it has been a significant period of time since I've done proofs, as statistics and probability classes don't use proofs! Also, I can't get \in to work. Sorry about that!

Homework Helper
Gold Member

## Homework Statement

If S1, S2 are nonempty subsets of ℝ that are bounded from above, prove that

l.u.b. {x+y : x \in S1, y \in S2 } = l.u.b. S1 + l.u.b. S2

## Homework Equations

Least Upper Bound Property

## The Attempt at a Solution

Using the least upper bound property, let us suppose that a is an upper bound for S1 and b is an upper bound for S2, as they are both a set of real numbers. Then there exists an x in S1 s.t. x≤a and there exists a y in S2 s.t. y≤b. By adding them together, x+y≤a+b where a and b are the upper bounds for their respected set of real numbers. Thus, it is equal to l.u.b. S1 + l.u.b. S2

I think this is how it goes. I'm sorry if my proof is horrible; it has been a significant period of time since I've done proofs, as statistics and probability classes don't use proofs! Also, I can't get \in to work. Sorry about that!

All that you have shown thus far is that if $x \in S_1$ and $y \in S_2$, then $x + y \leq lub(S_1) + lub(S_2)$.

This implies that $lub(x + y : x\in S_1, y\in S_2) \leq lub(S_1) + lub(S_2)$.

However, you still have to prove the opposite inequality.

utstatistics
Wait, sorry if I don't understand. Are you saying I have to prove the inequality when instead of it being ≤, I prove ≥?

Homework Helper
Gold Member
Wait, sorry if I don't understand. Are you saying I have to prove the inequality when instead of it being ≤, I prove ≥?

Yes. If your goal is to prove that A = B, and you have only proved A <= B, then you also need to show that A >= B before you can conclude that A = B.

utstatistics
...oh. WELL, excuse me while I think about this some more and get back to you. Haha.

utstatistics
So let me get this straight. I now have to prove that x ≥ a in S1 and y ≥ b in S2. BUT, isn't that a contradiction? If you add them, you get x+y≥a+b, but if the elements of x, y in their respective sets are greater than the l.u.b., then a and b are not the l.u.b.

I don't quite understand. Perhaps you can guide me in the right direction. Much thanks.

-J

Homework Helper
Gold Member
So let me get this straight. I now have to prove that x ≥ a in S1 and y ≥ b in S2. BUT, isn't that a contradiction? If you add them, you get x+y≥a+b, but if the elements of x, y in their respective sets are greater than the l.u.b., then a and b are not the l.u.b.

I don't quite understand. Perhaps you can guide me in the right direction. Much thanks.

-J

No, you have to show that $lub(x + y : x \in S_1, y \in S_2) \geq lub(S_1) + lub(S_2)$

I'm going to introduce some shorthand so it's easier to type:

a = lub(S_1)
b = lub(S_2)
c = lub(x+y : x in S_1, y in S_2)

If there were specific elements x in S_1 and y in S_2 such that x + y >= a + b, then you'd be done, because x + y <= c. But there might not be such elements. In fact, if there were, then you would have x = a and y = b. Thus each of the two sets S_1 and S_2 would contain its supremum. This is not guaranteed.

However, if you let $\epsilon > 0$, can you find x in S_1 and y in S_2 such that $x > a - \epsilon$ and $y > b - \epsilon$? i.e. you might not be able to find elements that achieve the upper bound, but can you get them arbitrarily close? If so, what can you do with that?

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