# Two small Analysis questions

1. Apr 18, 2013

### CAF123

1. The problem statement, all variables and given/known data
Prove:
1) a)If A is a nonempty bounded subset of $\mathbb{R}$ and B={$\epsilon x: x \in A$} then $\sup{B} = \epsilon \sup(A),$ where $\epsilon > 0$ any positive real number.

b) If A+B = {a+b, $a \in A, b \in B$}, A, B nonempty bounded subsets of $\mathbb{R}$ then $\sup(A+B) = \sup(A) + \sup(B)$

2)If $x_n$ converges, then $x_n/n$ converges.

3. The attempt at a solution

1) a) By Completeness axiom, $\sup(A)$ exists. Each element of B looks like $\epsilon x_i$, each $x_i \in A$. Write B = $\epsilon${x : $x \in A$} = $\epsilon A$. The supremum of A is an element of A and so let $\sup(A) = x_o \in A$. So $\sup(B) = \epsilon x_o = \epsilon (x_o) = \epsilon (sup(A))$.

b)$\sup(A)$ and $\sup(B)$ exist by Completeness. By defintion, $a \leq \sup(A) \forall a \in A,$ and $b \leq \sup(B) \forall b \in B$. So the largest element $a+b \leq \sup(A) + \sup(B)$ and the equality true if $\sup(A) \in A, \sup(B) \in B$. A + B is closed so the equality is true (not sure about this).

2) $x_n$ converges so $|x_n| \leq a \Rightarrow -a < x_n < a \Rightarrow -a/n < x_n/n < a/n$ i.e $|x_n/n| \leq a/n$. This does not prove anything helpful.

Instead, given that $x_n$ converges, we have that for $n \geq N, |x_n| \leq a$ so $1/n \leq 1/N \Rightarrow x_n/n \leq x_n/N \leq a/N$ I think this is wrong though since I have assumed the $x_i$ are positive. Is there a way to do this via epsilon-N? I can't see it yet.

Many thanks.

2. Apr 18, 2013

### LCKurtz

Why doesn't it? What happens as $n\rightarrow\infty$ in that last inequality?

3. Apr 18, 2013

### CAF123

Hi LCKurtz,
I don't know whether you can say anything as $n \rightarrow \infty$: All I have proven is that the sequence is bounded. But bounded $\neq$ convergence. E.g my sequence might fluctuate between -a/n and a/n indefinitely.

4. Apr 18, 2013

### LCKurtz

You have $\left|\frac{x_n} n\right|\le\frac a n$. $a$ is a fixed number. What happens to the right side of that as $n\rightarrow\infty$?

5. Apr 18, 2013

### CAF123

I see, I neglected that n wasn't constant. The RHS tends to zero, so that $x_n/n \rightarrow 0?$

6. Apr 18, 2013

### LCKurtz

Yes.

7. Apr 18, 2013

### jmjlt88

"The supremum of A is an element of A..."

The supremum of A may or may not be an element of A. For example, what is the sup A if A=(1,2)? What about if A=[1,2]? If we let, say, M= sup A, then one part of our defintion says that x≤M for all x in A. Since ε is a small positive number, εx≤εM for all x in A; this implies that εM is a upperbound for the set B.... Now, use the other part of our definition for the supremum of a set to finish it off!! =)

8. Apr 19, 2013

### CAF123

We can also write $\epsilon x \leq \sup(B)$. Also$\epsilon x \leq \epsilon \sup(A)$ also. We may have $\epsilon \sup(A) \leq \sup(B)$ or $\sup(B) \leq \epsilon \sup(A)$ in which case the equality holds.

Ideas for 1 b)? I said before that A + B was finite, but I don't think this is the case since A and B need not be finite. The only result I think I am getting is that the inequality holds: i.e since $a \leq \sup(A)$ and $b \leq \sup(B)$ then $a+b \leq \sup(A) + \sup(B)$ If $\sup(A) \in A,\,\sup(B) \in B$ then obviously the equality holds but that assumption is not applicable here.

9. Apr 19, 2013

### jmjlt88

"We can also write ϵx≤sup(B) . Alsoϵx≤ϵsup(A) also. We may have ϵsup(A)≤sup(B) or sup(B)≤ϵsup(A) in which case the equality holds."

I am not sure what you are getting at in the above statement. Read carefully the definition of the supremum of a subset of the real numbers. In my previous post, we let M= sup A. We desire to show that εM is the supremum of the set B. We have already shown that εM is an upperbound for the set B. What remains to be shown is that εM is the LEAST upper bound for the set B. Then, we can conclude
ε sup A = εM = sup B​
as desired.

Glancing at my analysis text (Ross), the author states that if A is a subset of the real numbers that is bounded above and M = sup A, then
(i) x≤M for all x in A, and
(ii) If M1< M, then there is an x in A such that M1<x.

We have already shown condition (i). Hence, we are left to verify that εM satisfies condition (ii). Here is a start: Let M0 < εM. Now, M0=εM1 for some real number M1.

From here, use the fact that M = sup A.

10. Apr 19, 2013

### CAF123

I don't have that definition in my book.. Anyway, that would imply $M_1< M = \sup(A)$. But we also know $x ≤ \sup(A)$ for x in A. So $M_1 < x$ since there exists an x such that supA - p≤ x ≤ sup A, p small > 0.

11. Apr 21, 2013

### jmjlt88

Right! Now, because
M1<x≤M​
we have that
______________ . ​

12. Apr 21, 2013

### CAF123

Then there exists an x greater than this M1 and smaller than sup A, so there is an εx greater than εM1 and smaller than εsup A. So from this, I think I can conclude that $\epsilon \sup A$ is the lowest upper bound and since $\epsilon x \leq \sup B$ also I can say $\epsilon \sup A = \sup B$. Right?

Last edited: Apr 21, 2013