# I Upper bound and supremum problem

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1. Apr 12, 2017

### i_hate_math

Claim: Let A be a non-empty subset of R+ = {x ∈ R : x > 0} which is bounded above, and let B = {x2 : x ∈ A}. Then sup(B) = sup(A)2.

a. Prove the claim.
b. Does the claim still hold if we replace R+ with R? Explain briefly.

So I have spent the past hours trying to prove this claim using the theorem:
a = sup(A) ⇔ ∀ ε>0 (( ∀ x∈A x ≤ a + ε )∧( ∃ x∈A x > a - ε ))

I aim to prove true for: ∀ ε>0 (( ∀ x∈A x2 ≤ a2 + ε )∧( ∃ x∈A x2 > a2 - ε )), which is equivalent to proving ∀ ε>0 (( ∀ x∈B x ≤ b + ε )∧( ∃ x∈B x > b - ε )), given how set B is related to A.

As for the proof:
Firstly, it is given that A is bounded above, this means sup(A) exists.
Letting a = sup(A) gives: ∀ ε>0 (( ∀ x∈A x ≤ a + ε )∧( ∃ x∈A x > a - ε )) statement 1
I was able to use the definition of supremum to get a≥x and since a,x>0 , a2≥x2. With this result, it is easy to get the first part of the definition for B: ∀ ε>0 ( ∀ x∈A x2 ≤ a2 + ε ).
But the this result is not as helpful when proving the second part (namely, ∀ ε>0 ( ∃ x∈A x > a - ε )), neither is statement 1 (tried squaring both sides and messing with the inequalities but didn't work).

What I really need to prove is just there exist an x such that x2 > a - ε, this seems almost to be the definition of supremum. I took a further step, attempting to prove by cases with case 1 being a∈A, this gives the desired result easily; case 2 is a∉A, again, the difficulties arise.

Last edited: Apr 12, 2017
2. Apr 12, 2017

### PeroK

If you can assume (or prove) that for $x, y > 0, \ x \le y \ \Leftrightarrow x^2 \le y^2$, then I would suggest a different approach using the definition of supremum as the least upper bound.

Last edited: Apr 12, 2017
3. Apr 12, 2017

### andrewkirk

The second part is readily proved by contradiction. Assume it is false, ie $\neg(\forall\varepsilon>0,\ \exists x\in B:\ x>a^2-\varepsilon)$, where $a\triangleq \sup(A)$. That is equivalent to $\exists \varepsilon>0,\ \forall x\in B:\ x\leq a^2-\varepsilon$ which, because the square function is a monotonic bijection between $A\subseteq\mathbb R^+$ and $B\subseteq\mathbb R^+$, is equivalent to $\exists \varepsilon>0,\ \forall y\in A:\ y^2\leq a^2-\varepsilon$.

Try completing the square of $a^2-\varepsilon$ in order to find an upper bound for $A$ that is less than $a$, which will give us the contradiction we need.

4. Apr 12, 2017

### i_hate_math

I don't quite get what you're suggesting. Could you be a bit more specific about using the definition of supremum?

5. Apr 12, 2017

### i_hate_math

Thank you very much for your quick response, very thorough explanation. I'll work on it!

6. Apr 12, 2017

### PeroK

The supremum is the least upper bound. $a = sup(A)$ means:

1) $a$ is an upper bound for A ($\forall x \in A, x \le a$)

2) If $b$ is an upper bound for $A$ then $a \le b$

That's what a supremum is, by definition.

To help you a little, if $a = sup(A)$ you could show thta $a^2 = sup(B)$ by showing that:

1) $\forall x \in B, x \le a^2$

2) If $b$ is an upper bound for $B$ then $a^2 \le b$

7. Apr 12, 2017

### i_hate_math

I am able to prove (1) as it is equivalent to ∀x∈A x2 ≤ a2, which can be derived from ∀x∈A x ≤ a (for positive x and a).
But I am having trouble with the second part, because the only thing we know about b so far is ∀x∈b x ≤ b, can also be written as ∀x∈A x2 ≤ b. This does not , though, give the desired a2≤b. Could you enlighten me a bit further?

8. Apr 12, 2017

### i_hate_math

I've tried your suggested method, and it worked well and I am well convinced. However, there are still a few things I am not certain:
1. Is it sound to get from ∀ε>0 ∃x∈A, x≤a that ∀ε>0 ∃x∈A, x2≤a2 ?
2. For when I am rewriting the logical notations of these statements, do I need to assign different parameters x, y .etc for elements in different sets? e.g. x and y in your reply

Thanks heaps.

9. Apr 12, 2017

### andrewkirk

Yes, it is sound because
$$x\geq 0\to\left(x\leq a\ \leftrightarrow x^2\leq a^2\right)$$
It would be worthwhile to prove that formally as an exercise.

It is the absence of that antecedent condition $x\geq 0$ that prevents part (b) of the question in the OP from being true.
It is not strictly necessary, because the scope of a variable is restricted to the smallest containing formula in which it is quantified. However I think it is good practice to use different variables for members of different sets, because it makes it easier for the reader to follow.

10. Apr 13, 2017

### PeroK

You need to use the fact that $a = Sup(A)$. Hint: use $\forall x \in A, x \le \sqrt{b}$