Prove that $\forall z \in(0,1)$, $\frac{z+1}{2}>z$

Question

For some context, the initial problem that I was working on was proving that for the set $A=(0,1)$ we have $\sup(A)=1$. My argument relied on the fact that for $z \in(0,1)$ we have that $$\frac{z+1}{2}>z $$ However, I wanted to prove rigorously that this result is true (it is given without proof in the lecture notes I am reading). Here is my attempt:

First I would like to prove that $\frac{z+1}{2} \in (0,1)$. We know that $z\in(0,1)$, thus $$0<z<1 $$ Which implies that $$0<\frac{z}{2}<\frac{1}{2} $$ Furthermore $$\frac{1}{2}<\frac{z+1}{2}<1$$ Hence it is clear that $\frac{z+1}{2} \in (0,1)$

I now aim to use contradiction to show that $$\frac{z+1}{2}\le z \text{ on } z\in(0,1)$$ is impossible. Cleary equality fails as the only solution is $z=1 \notin (0,1)$, so we are left to show that $\frac{z+1}{2}<z$ is false.
By assumption we have $$0<\frac{z+1}{2}<z $$ We may rearrange this to give $$0<z+1<2z $$ Furthermore $$-1<z<2z-1 $$ Finally $$-1-z<0<z-1 $$ Clearly this gives that $0<z-1$ or in other words, $1<z$ which is equivalent to $z>1$ which can't be as $z\in(0,1)$. The negation of our original statement is false, and so the original statement must be true.

Is my reasoning correct?

Answer 1

This works, but there is no need for a contradiction: a direct proof is a single line.

Since $z \in (0,1)$, then $1>z$. Therefore, $$ \frac{z+1}{2} > \frac{z+z}{2} = z $$

Answer 2

Another way to approach the supremum problem. By definition, the supremum of a given set $A \neq \emptyset$ is the minimum element which is bigger or equal than every element of $A$. In other words, if we consider the sets of upper bounds of $A$, let's denote it $\mathcal U$, then $$\sup(A) = \min (\mathcal U)$$ Note that it always exists. If $A=(0,1)$, then $$\mathcal U = [1,+\infty)$$ Therefore. $$\sup(A) = \min (\mathcal U) = \min \{x\in [1,+\infty)\} = 1$$