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

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.
