# Nonempty subset homework

1. Apr 27, 2013

### Zondrina

1. The problem statement, all variables and given/known data

http://gyazo.com/27a104e05c409c75611ee4250a89c790

2. Relevant equations

Sup/Inf axioms as well as the ε-N definition.

3. The attempt at a solution

Suppose A is a nonempty subset of ℝ bounded above by $M$. We want to show that $lim_{n→∞} x_n = sup(A)$ where $x_n$ is a sequence of elements of A.

That is, $\forall ε > 0, \exists N \space | \space n ≥ N \Rightarrow |x_n - sup(A)| < ε$

Note that since A is bounded above by M, we know that $M > a, \space \forall a \in A$ including $sup(A)$.

So :

$|x_n - sup(A)| ≤ |x_n| + |sup(A)| < |x_n| + M$

So $|x_n|$ is bounded above by $ε - M$ and below by $-(ε+M)$

I'm not quite sure how to continue here.

2. Apr 27, 2013

### micromass

You need to define the sequence $(x_n)_n$ first. Of course not every sequence will do.

3. Apr 27, 2013

### Zondrina

Ohhh I thought this was a general question, so I need a concrete sequence then?

How about $A = \{2 - \frac{1}{n} \space | \space n = 1,2,... \}$ and $x_n = 2 - \frac{1}{n}$.

Then $sup(A) = 2$ and $x_n → 2$ as $n→∞$

$|x_n - sup(A)| = |2 - \frac{1}{n} - 2| = \frac{1}{n}$

So choosing $n ≥ \frac{1}{ε} \Rightarrow |x_n - sup(A)| < ε$

Didn't even need the upper bound except to establish sup(A) existed :).

4. Apr 27, 2013

### micromass

You'll need a concrete sequence. But the set $A$ is completely general. You can't pick $A=\{2-1/n~\vert~n\in \mathbb{N}\}$. It needs to work for every possible set $A$ (that is bounded above).

So for a set $A$ that is bounded above, you need to find a concrete sequence that converges to $sup(A)$.

5. Apr 27, 2013

### Zondrina

I'm having a bit of trouble actually producing the sequence. I understand why I cannot pre-define A ( I did it mostly to get an idea of whats going on ), but can I choose ANY convergent sequence or is there a method to doing this?

I mean I can come up with enough convergent sequences like $x_n = 1/n$, but it's not like I can put a finite number on sup(A).

All i really have is the fact that A is bounded above by M so that sup(A) exists.

6. Apr 27, 2013

### micromass

Here's a hint. Prove the following "lemma":

Given a positive integer $n$. There is always an element $x_n$ such that $sup(A)-\frac{1}{n}< x_n$ and such that $x_n\in A$.

7. Apr 27, 2013

### Zondrina

I don't see how that helps me? I've been staring at it for awhile and I'm unclear about your intentions.

sup(A) < xn + 1/n

Then i was thinking induction.

8. Apr 27, 2013

### micromass

Induction won't help here.

The intention of my lemma is that it would define a nice sequence $(x_n)_n$. The sequence as defined in the lemma can be shown to converge to $sup(A)$.

9. Apr 27, 2013

### Zondrina

Waaaait I see what you're getting at now. Since $x_n + 1/n$ will be a sequence as well.

I unfortunately just got called into work 2 mins ago, I will continue this later today when I get home.

10. Apr 27, 2013

### micromass

It will be a sequence as well, but I'm not sure how it helps...

Have fun!

11. Apr 28, 2013

### Zondrina

Let $A$ be a nonempty subset of $ℝ$ which is bounded above.

We must show $\exists x_n \in A \space | \space \lim_{n→∞} x_n = sup(A)$

We must somehow construct $x_n$ from the elements of A so that it converges to sup(A).

Suppose n is a positive integer and $x_n$ is a sequence of elements of A. Since $x_n \in A$, we know that $x_n ≤ sup(A)$ so that $x_n$ is bounded.

I think I have to use the fact it's bounded somehow, I'm having some thoughts, but I'd rather hear some input before going off in potentially the wrong direction.

12. Apr 28, 2013

### Zondrina

Sorry for the double, but I just woke up so I want to put a fresh brain effort into this.

Let $A$ be a nonempty subset of $ℝ$ which is bounded above.

We must show $\exists x_n \in A \space | \space \lim_{n→∞} x_n = sup(A)$

We must somehow construct $x_n$ from the elements of A so that it converges to sup(A). Now because $x_n \in A$, we know that $x_n ≤ sup(A)$.

Using the hint given, I want to show the inequality $x_n ≥ sup(A) - \frac{1}{n}$ holds for any positive integer n because $sup(A) - \frac{1}{n}$ is not an upper bound for A. Inductively, take n=1. Then $x_1 ≥ sup(A) - 1$ and now assume this holds for any n. We want to show this holds for n+1, so $x_{n+1} ≥ sup(A) - \frac{1}{n+1}$.

So we can inductively build our sequence $x_n$ with following the relation $x_n ≥ sup(A) - \frac{1}{n}$.

Now notice that :
$x_n ≥ sup(A) - \frac{1}{n}$
$\Rightarrow sup(A) - x_n ≤ \frac{1}{n}$
$\Rightarrow |x_n - sup(A)| ≤ \frac{1}{n}$

So, $\forall ε>0, \exists N \space | \space n ≥ \frac{1}{ε} \Rightarrow |x_n - sup(A)| < ε$

Thus making our choice of $n≥\frac{1}{ε}$ we observe $|x_n - sup(A)| ≤ \frac{1}{n} < ε$.

Therefore we have found a way to construct a sequence of elements of A which converges to sup(A) as desired.

I know you said not to use induction micro, but I didn't seem to see any other way. Hopefully this makes sense.

13. Apr 28, 2013

### christoff

In the above, you haven't actually shown that $x_n\in A$; you seem to be assuming this. Specifically, you need to show there exists a sequence that satisfies the inequality you want. Specifically, you need that for every positive integer, there exists an $x_n \in A$ such that $sup(A)<x_n+1/n.$ I would refer you to another problem you are working on that shows you that such an $x_n$ always exists..

ie. show that s=sup(A) if and only if a≤s for all a in A, and for all $\epsilon>0$, there exists $a\in A$ such that $s-\epsilon<a$.

Spoiler below (try working it out on your own first).

To get $x_n=a$, try taking $1/n=\epsilon$.

Last edited: Apr 28, 2013
14. Apr 29, 2013

### Zondrina

Yes my wording was wrong there I meant to say because sup(A) exists, let me try this proof again.

Let $A$ be a nonempty subset of $ℝ$ which is bounded above.

We must show $\exists x_n \in A \space | \space \lim_{n→∞} x_n = sup(A)$

We must somehow construct $x_n$ from the elements of A so that it converges to sup(A), but because sup(A) exists we know that $x_n ≤ sup(A)$

So for any positive integer n, we must show $sup(A) - \frac{1}{n} < x_n ≤ sup(A)$.

This comment made me blank for awhile :

ie. show that s=sup(A) if and only if a≤s for all a in A, and for all ϵ>0, there exists a∈A such that s−ϵ<a.

I don't see how that relates to this problem?

15. Apr 29, 2013

### christoff

For n=1, you have $\epsilon=1/1=1$, so there exists an $x\in A$ such that
$sup(A)-1<x$. Denote this x by $x_1$.

For n=2, you have $\epsilon=1/2$, so there exists an $x\in A$ such that $sup(A)-1/2<x$. Now denote THIS x by $x_2$.

For n=3, $\epsilon=1/3$. Construct $x_3$ the same way.

The existence of each $x_n$ at each step is guaranteed by the other problem I cited, which you just finished in another thread. In particular, such an $x_n$ exists for each n, and satisfies... which inequality?

As for why it's a sequence... to each n, you can associate an $x_n$, and each of these is in $A$..

Last edited: Apr 29, 2013
16. Apr 29, 2013

### Zondrina

Ohh I see. Let me try again then.

Let $A$ be a nonempty subset of $ℝ$ which is bounded above.

We must show $\exists x_n \in A \space | \space \lim_{n→∞} x_n = sup(A)$

So for any positive integer n, we must show $sup(A) - \frac{1}{n} < x_n$ so we can construct a sequence which is bounded by $sup(A)$.

So take n=1, then we can find $x_1 \in A$ such that $sup(A) - 1 < x_1$ because sup(A)-1 is not an upper bound for A. We can follow this construction for any positive n yielding our desired inequality.

So we have successfully constructed our sequence $x_n$ by choosing the elements of A which satisfy $sup(A) - \frac{1}{n} < x_n$. Now because sup(A) exists, we know that $a ≤ sup(A), \space \forall a \in A$. Since each $x_n \in A$ for each positive integer n, we know that $x_n ≤ sup(A)$ for all n so that the sequence $x_n$ is bounded above by sup(A).

Hmm I don't have access to the monotone convergence theorem until literally one question after this. How would I finish this up without it? The same way I did before a few posts ago?

Last edited: Apr 29, 2013
17. Apr 29, 2013

### CAF123

I think we are working on the same (or similar question). I just said since you have sup A -1/n < x ≤ sup A and then when you take the limit the left most inequality necessarily becomes a less or equal. So x is less or equal to sup A on both sides which means it is equal to sup A.

18. Apr 29, 2013

### Zondrina

Yes I was thinking about this, but I wasn't sure if I could take limits for granted. That would indeed show that $lim(sup(A) - \frac{1}{n}) < lim(x_n) ≤ lim(sup(A))$, but because we can't have sup(A) < sup(A), it must be the case that $lim(x_n) = sup(A)$.

19. Apr 29, 2013

### christoff

How to bypass this thing with the inequalities;
Note that if $a<b$, the we also have $a\leq b$. So whenever you have a less-than sign, you can always instead write a less-than-or-equal-to sign.
Example: we have 1<2. Also, 1≤2.
On the other hand, the opposite is false; you can't necessarily turn a ≤ into a <.
Example: we have 2≤2. We don't have 2<2.

As for the aspect of "taking the limits for granted", here's a way to do it rigorously:

For all n>0, we have $x_n-sup(A)\leq 0$. On the other hand, we have $-\frac{1}{n}\leq x_n-sup(A)$ for all n. Putting these two together, we have $-\frac{1}{n}\leq x_n-sup(A)\leq 0 \leq \frac{1}{n}$. Hence, $|x_n-\sup(A)|\leq\frac{1}{n}<\frac{2}{n}$.

Now let $\epsilon>0$. Then there exists an $N>0$ such that $\frac{2}{N}<\epsilon$ (just take N large enough; $N>\frac{2}{\epsilon}$). Then for all $n>N$, we have $|x_n-\sup(A)|<\frac{2}{n}<\frac{2}{N}=\epsilon$. Hence, $x_n$ converges to $sup(A)$ (by definition of convergence; epsilon definition).