Proof of contradiction when showing the supremum of a function 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

"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.
=====
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)
