Prove that there exists $b \in B$ that is an upper bound for $A$

Question

Given that $ \ Sup A < \ Sup B$

Now $\forall \epsilon, \exists b \in B \ such \ that $

$ \ Sup A - \epsilon < \ Sup B - \epsilon < b$

SO,

$ \ Sup A - \epsilon < b $

So we have now

$\forall \epsilon, \exists b \in B \ such \ that $

$ \ Sup A - \epsilon < b $

Also $ \forall a \in A, a \leq \ SupA. \ So, a - \epsilon \leq \ SupA - \epsilon $

So we have $ \forall a \in A, \exists b \in B \ and \forall \epsilon >0$

$a - \epsilon < b $

$a < b- \epsilon$

How do i procced ?

Thanks

Answer 1

Think of a strong statement as one that applies in many cases, and a weak statement as one that does not apply in too many cases. For example, the statement : "every odd number is not a multiple of $2$" is stronger than the statement : every power of an odd prime is not a multiple of $2$, which is stronger than the statement : a power of three is never a multiple of $2$ , which is stronger than the statement : $27$ is not a multiple of $2$.

Now, imagine I gave you the first statement. It has a lot of information, since it's saying something for each odd number. After that, we are going down to smaller and smaller classes of odd numbers, till we reach a statement which is so easy to verify and contains information only about one number, $27$.

Now, the problem with proceeding like this, is that while we have a lot of true statements with us, as we have proceeded we have weakened each statement, so that the last statement is something which you really cannot do anything with.

Something like this has happened in your working. Let me explain.

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Now, you started off with $\sup A < \sup B$.

Now, $\forall \epsilon \exists b \in B$, such that $\sup A - \epsilon < \sup B - \epsilon < B$.

Fine, this is by definition, so it is just a rewrite and using the fact that $\sup A < \sup B$.

So, $\mathit{\sup A - \epsilon < b}$.

This is a weakening of the previous line. The point is that you have now left out the fact that $\sup B - \epsilon$ lies in between these (sure, it is bigger than $\sup A - \epsilon$, but you have now forgotten that $b$ is bigger than it, or are not planning to use that information, which could be key in solving the problem).

Also, $\forall a \in A, a \leq \sup A$ so $a - \epsilon \leq \sup A - \epsilon$.

Once again, a correct statement, but one drifting away from the problem since to show that any $b$ is an upper bound of $A$ you only need to show that $b \geq \sup A$, so you don't need to involve individual elements of $a$, which unfortunately is being done here.

So we have $\forall a \in A$ there is $b \in B$ such that $\forall \epsilon >0$ , $a \leq b - \epsilon$.

The problem with this statement, is that again while it is correct, you actually are saying that for each $a$ there is a $b$ depending on $a$ so that for each $\epsilon$ we have $a-\epsilon < b$. Now, first of all the $b$ depends on $a$, so how are you going to pick one of these $b$s to satisfy the solution? Next, $\sup A$ is part of the conclusion (as I mentioned, it is enough to find $b \geq \sup A$) but is not involved in this reasoning. Therefore, this weakest statement of your reasoning, while being correct, does not seemingly have the required information to achieve the conclusion.

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That is, if you want to rescue the conclusion from your last line of working ,you are now left with the following question :

Suppose $A,B$ are sets such that for all $a \in A$ there is $b \in B$ such that for every $\epsilon > 0$ we have $a -\epsilon< b$. Is it true that some $b \in B$ is an upper bound of $A$?

Now, it turns out that this is false! First of all, note that the "$\forall \epsilon > 0$ we have $a-\epsilon < b$" is the same as $a \leq b$. The counterexample is just $A = B = [0,1)$, since here one sees that no element of $B$ is an upper bound of $A$, since no element of $B$ is bigger than $\sup A = 1$. But $A,B$ satisfy the hypothesis of this question.

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Which is the classical situation of overworking : you reach a very weak statement which does not imply the conclusion in general. Then, you have to work back carefully and find out what things you neglected : where did you weaken your hypothesis so much that the conclusion could not be derived.

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That is what I pointed out earlier : neglecting $b > \sup B - \epsilon$ was crucial to the proof. Without that, you cannot really proceed.

So, in your working, you have to (unfortunately) scratch out everything after $\sup A - \epsilon < \sup B - \epsilon < b$.

Now, you have to think : I want a single $b$. To get a single $b$, I need a specific value of $\epsilon$. Which one can I use?

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The point is, that imagine a seemingly very small change in your question : given $\sup A \color{blue}{\leq} \sup B$, prove that some $b \in B$ is an upper bound of $A$. This statement is false with the same $A,B$ which I used earlier.

Therefore, we have to use the fact that $\sup A \color{green}{<} \sup B$. The fact that the separation $\sup B - \sup A$ is strictly positive hass to be used for the question. Let us call $\sup B - \sup A = \delta > 0$.

Now, the idea is this : if $L = \sup A$, then all we need to show that there is some $b \in B$ with $b \geq L$. However, note that if $(L+\delta) - b < \delta$ then rearranging gives us $b > L$.

But $(L+\delta) = \sup B$, so we are essentially saying that if $\sup B - b < \delta$ then $b > L$. Can we find such a $b$? Yes, from the property of supremum and the fact that $\color{orange}{\delta > 0}$.

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Now, finally let me propose a proof:

Let $\sup B - \sup A = \delta > 0$. By the property of supremum there is a $b \in B$ such that $\sup B - b < \delta$. Rearrange to get $b > \sup B - \delta = \sup A$, so that $b > \sup A$ and hence $b$ is an upper bound of $A$.

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Even when you are getting correct statements, you should be careful that you do not encounter yourself in the presence of a very weak statement. This will force you to go back to an earlier statement. So, try to explore your options in full generality : a chase for "correct statements" is ok, but you must cut to the chase of the answer, for that is your aim. This will come both with practice and with exposure to the specific topic and its techniques.

Answer 2

Assume $\sup < \sup $

$\exists b\in B$ st. $ \sup < b < \sup B$ and thus $\sup < b$

(Because if there's no such $b \in B$, $\sup A$ would be an upper bound of B, contradiction.)

Answer 3

Attempt:

0) Given: $\sup A < \sup B$

1) $\sup A$ is an upper bound for $A$, the least upper bound, hence:

For $a \in A$: $a \le \sup A$ ,

2) Since $ \sup A < \sup B$ , $\sup B$ is an upper bound for $A$:

For $a \in A$: $a < \sup B$.

3)Since $ \sup B$ is the least upper bound for $B$:

For $b \in B$: $b \le \sup B$.

4) Set $d:= (1/2)(\sup B- \sup A) >0.$

A bit trickier now:

5) Since $\sup B$ is the the least upper bound for $B$:

For every $\epsilon > 0$:

$\sup B - \epsilon$ is NOT an upper bound for $B$.

Hence there is a $b_{\epsilon} \in B$ s.t.

$\sup B - \epsilon < b_{\epsilon} \le \sup B.$

(Note: If the above statement were not true $\sup B -\epsilon$ would be a bound for $B$, less than $\sup B$, a contradiction)

6) Let $\epsilon = d;$

Then

$\sup A < \sup B - d=$

$ (1/2)(\sup A +\sup B) < b_{d}$ .

Hence $b_{d}$ is an upper bound for $A$.