Proof of contradiction when showing the supremum of a function

Question

Say you wanted to show

$\sup{A}=4$

for every

$a_{n}=\frac{4n}{n+1}, n\in \mathbb{N}$

by proof of contradiction.

First, note that

$a_{n}=\frac{4n}{n+1} < \frac{4n}{n} = 4$

And we assume it is $\sup{A}$

Then, you may simplify the expression of $a_n$

$a_{n}=\frac{4n}{n+1}=4 - \frac{4}{n+1}$

Now we're looking to show that 4 is in fact the supremum of A. Doing so with proof of contradiction, we assume there is a supremum

$4 - \varepsilon, \varepsilon > 0$

Look for an n such that

$a_{n}=4-\frac{4}{n+1} > 4 - \varepsilon$

And then you solve for $n$. But apparently it means for all $n$ larger than whatever is on the other side, there is a number larger than if the supremum was $4-\varepsilon$. Now, what if you first assume the supremum of A is 5, and you do the same process? There is nothing apparent in the solution that makes the answer seem wrong! So the question is how does the proof work, if numerically it makes sense for other values for the supremum? Thanks

Answer 1

You can immediately rule out any real number larger than $4$ as the supremum of your sequence once you've rewritten the terms of your sequence in the form $a_n = 4 - \frac{4}{n + 1}$. It is now clear that all the terms of your sequence are strictly less than $4$, so $4$ is an upper bound for the sequence. By definition of the supremum, it follows that no real number larger than 4 could be the supremum of this sequence. So, the proof by contradiction you propose really just needs to show that assuming the supremum of the sequence is less than 4 yields a contradiction.

Answer 2

You showed that $4$ is an upper bound. Next you need to show it is the least upper bound. That means you need to show that for any $\epsilon > 0$, $4 - \epsilon$ is not an upper bound. And to show this all you need is finding a number $n$ such that $a_n > 4 - \epsilon$. So we solve: $\dfrac{4n}{n+1} > 4 - \epsilon \iff 4n > 4(n+1) - (n+1)\epsilon \iff (n+1)\epsilon > 4\iff n\epsilon > 4 - \epsilon \iff n > \dfrac{4 - \epsilon}{\epsilon}$. Thus you can choose $n$ such that $n > \dfrac{4-\epsilon}{\epsilon}$.

Answer 3

"Now, what if you first assume the supremum of A is 5, and you do the same process?"

But you can't do the same process.

You have shown that for all $a_n$ that $a_n < 4 < 5$ so if you try to solve $4 < 5-\epsilon < a_n < 5$ for some $\epsilon: 0 < \epsilon < 1$ you WILL fail. There is not any $a_n$ so that $4 < 5-\epsilon <a_n < 5$.

Let's see how we fail and what will happen if we try the same process:

Let $\epsilon > 0$.

Let $a_n = \frac 4{n+1} > 5 -\epsilon$ solve for $n$.

$\frac 4{n+1}= 4- \frac 4{n+1} > 5 -\epsilon$

$\epsilon -1 > \frac 4{n-1}$

$(n+1)[\epsilon - 1] > 4$. Now we need this to be true for every possible $\epsilon > 0$ so it must have a solution for $\epsilon = 1$ and it must have a solution for any $\epsilon < 1$.

If $\epsilon = 1$ we have $(n+1)\times 0 > 4$ or $0 > 4$ which has no solution. And if we have $\epsilon < 1$ we have $\epsilon - 1 < 0$ so

$(n+1)[\epsilon - 1] > 4$

$n+1 < \frac 4{\epsilon -1}$. But $\frac 4{\epsilon -1}$ is a negative number!

So we have $n + 1 < \frac 4{\epsilon -1} < 0$ and $n < -1$. But that contradicts $n\in \mathbb N$.

......

Basically if $\sup A = 4$ then $\sup A < 4=\sup A$ is a contradiction. But $4= \sup A < 5$ is not a contradiction.

......

Basically we need two things for the definition of $s = \sup A$.

We need

that $s$ is an upper bound of $A$.

That's fine $4$ is an upper bound. So is $4.1$ and so is $5$ and so is $27$ and so is $3$ billion.

But we also need

there is no upper bound that is less than $s$.

That is true for $s = 4$. And we can prove that by solving for $a_n = \frac 4{n+1} > 4-\epsilon$ always has a solution (which means $4-\epsilon$ can not be an upper bound.

But it is not true for $4.1$ or $5$ or $27$ or $3$ billion because $4 < 4.1 < 5 < 27 < 3$billion and $4$ IS an upper bound.

And we tried to prove that $4$ or $4.1$ or $4.999999$ are not upper bounds we will fail because ... they are upper bounds.

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For what it's worth:

I dislike the idea of "assume $\sup = 4$" and then "assume there is a $\sup A < 4$".

It's better to show $4$ is an upper bound, then assume there is an upper bound (maybe the $\sup$, maybe not) that is less than $4$. You show that is a contradiction but showing for ALL $\epsilon > 0$ then there will exist an $a_n$ so that $4 - \epsilon < a_n\in A$ (so $4-\epsilon$ can not be an upper bound)