Q:

Formally Proving $1$ is the supremum of $(0, 1)$ in $\mathbb{R}$

I'm just getting used to doing very formal proofs in real analysis, and this problem has stumped me in terms of formulating a proof:

Prove, or disprove, that $1$ is the supremum of $(0, 1)$ in $\mathbb{R}$.

It's intuitively clear that $1$ is the least upper bound of $A = (0, 1)$; $1$ is at the very "edge" of this interval, and even if you go an infinitesimally small amount below $1$, it's still inside $A$ and less than $1,$ which is sup $A$. But, how do I prove this formally? I tried using an epsilon definition, but I seem to be going in circles. Any help would be great.
Also, how does one become better at doing these "simple" but formal proofs? I seem to struggle with proving things as seemingly easy as "prove that $a * 0$ = $0$ in a field". Thank you.

A:

For each $x\in(0,1)$, $x\leqslant1$. Therefore, $1$ is an upper bound of $(0,1)$.
And if $x<1$, if $x<0$, then $x$ is not an upper bound of $(0,1)$. Otherwise, $x<\frac{x+1}2\in(0,1)$, which, again, proves that $x$ is not an upper bound of $(0,1)$.
So, $1$ is the least upper bound of $(0,1)$.

A:

Hint: $s$ is the supremum of some set $A$ if for each $\epsilon>0$ there exists some $a \in A$ so that $s-a<\epsilon$.
Specializing: let $s=1$, and consider $1-\frac{1}{n} \in  (0,1)$ for each $ n \in \mathbb N$.

