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

Question

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.

Answer 1

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 $(0,1)$.

Answer 2

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$.

Answer 3

As Andres Mejia said, consider que set $A=\{x\in\mathbb{R}:0<x<1\}$.

By definition, $\alpha =Sup(A)$ if and only if

1. $\alpha$ is the upper bound of $A$ and
2. $\forall \epsilon >0 \exists a_0\in A (\alpha-\epsilon<a_0)$.

By definition of set $A$, $\forall x\in A, x<1$, so $1$ es an upper bound of $A$. We need now to prove 2). Let $\alpha = 1$. Suppose that $\epsilon >0$. We want to find $a_0 \in A$ such that $\alpha-\epsilon<a_0$. This means that $1-\epsilon<a_0$.

Thus $\epsilon>0$, by the archimedean property, $\exists n_0\in\mathbb{N}$ such that $\frac{1}{n_0}<\epsilon$ therefore, $-\epsilon < -\frac{1}{n_0}$ and therefore $1-\epsilon < 1-\frac{1}{n_0}$, which means that $a_0=1-\frac{1}{n_0}$ just as Andres said.

$\therefore Sup(A)=1$.

Same thing happens with infimum.