Proving Properties of Bounded Sets in Real Numbers

[CAF123]

Homework Statement

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.

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.

 

[LCKurtz]

Homework Statement

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

The Attempt at a Solution

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.

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

 

[CAF123]

Hi LCKurtz,

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

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.

 

[LCKurtz]

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.

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 $n\rightarrow\infty$?

 

[CAF123]

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 $n\rightarrow\infty$?

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

 

[LCKurtz]

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

Yes.

 

[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! =)

 

[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.

 

[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.

 

[CAF123]

"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.

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.

 

[jmjlt88]

Right! Now, because

M₁<x≤M​

we have that

______________ .​

 

[CAF123]

Right! Now, because

M₁<x≤M​

we have that

______________ .​

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?