Proving rigorously the supremum of a set

I think there's a way more simple and intuitive proof.

First, as you observed, it is obvious that 2 is an upper bound. Now, to prove it is the supremum. Assume that $M$ is the supremum and $M<2$.

Of course, $M>2$ is trivially impossible since $2$ is an upper bound as well and thus any $M$ bigger than $2$ cannot be a supremum.

Now, let $x=\frac{2-M}{2}+M$.

As you can obviously see $M<x$, so all that is left is to show that $x<2$.

But now, assume it is not, that is $x\geq2$.

Then, $\frac{2-M}{2}+M\geq2$. Multiplying both sides by $2$, we get

$2-M+2M\geq4$, that is $2+M\geq4$. But $M<2$, so $2+M<4$. Thus, by reductio ad absurdum, $x<2$. This shows that $M<2$ cannot be the supremum.

QED.

Now, as for the simplicity of this proof, I have written a lot for clarity and in case you are a beginner on this subject. This can be summarized in $2$ lines, but this is for clarity. I hope this helps, and you must soon learn to find the shortest and more intuitive way. Good luck.

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Let $a\lt 2$ is $\sup A$.

Therefore, $a=2-b$ , where $b\gt 0$.

We can get some $n\in \mathbb{N} \,|\, 0 \lt \frac1n \lt b$

So, $2-b\lt2-(\frac1n)\lt2$.

$\exists c\in Q\cap A\,|\,2\gt c\gt 2-(\frac1n)\gt 0$.

So,$\,2\gt c \gt 2-(\frac1n)\gt (2-b)=a$.

$0\lt c \lt 2 \implies c\in A$, and also $c\gt a$

i.e. $a$ is not even an upper bound of $A$.

$\implies$ the set of upper bounds of $A$ is $\{x : x\in \mathbb{R}$ and $x\ge 2\}=S$ and the least member of $S$ is $2$.

So, $2$ is the least upper bound of $A$.