Formal proof of infimum of $(0,1)$ I have to prove that infimum of $(0,1)$ is $0$.
Let $A=(0,1)=\{x:0<x<1\}$. clearly $0$ is lower bound of $A$. Let $a$ be a lower of $A$. Since $\frac{1}{2}\in A$, we have $a\leq \frac{1}{2}<1$. We claim that $0\geq a$. If not, $a>0$, then $a\in A$ and $\frac{a}{2}\in A$ then $a<\frac{a}{2}$. contradiction! Since $a$ is lower bound of $A$.(edited.)

But I didn't find contradiction.
I don't know where  I am wrong? I tried  a lot but i didn't prove it completely.
Any help will be appreciated. thanks!
 A: What you have to show:
Step 1: $0$ is a lower bound for $(0,1)$. Done; trivial as $x \in (0,1)$ implies $x > 0$ so $x \ge 0$ a fortiori.
Step 2: if $a$ is any other lower bound, $a \le 0$.
So suppose for a contradiction that $a$ is a lower bound for $(0,1)$ and $a>0$. Then $\frac{a}{2} \in (0,1)$ and $a \not \le \frac{a}{2}$ so contradiction as $a$ is not a lower bound for $(0,1)$ after all.
So $a \le 0$ from $\lnot (a >0)$.
A: First, $0$ is a lower bound. Moreover, if $\varepsilon \in (0,1)$, then $x:=\frac{\varepsilon }{2}\in (0,1)$ and $$x\leq 0+\varepsilon,$$
The claim follow.
A: It is clear that $0$ is the lower bound of $A=(0,1)$, as $a\gt 0 , \forall a \in A$
We should prove that , if $x $ is a lower bound of $A$ other than $0$ ,then, $x\lt 0 $.
If possible, let , there is lower bound $b$ of $A$ such that $b\gt 0 $ , then we choose a $\delta\gt 0 $ such that $b-\delta \gt 0 $, this is a contradiction to the fact that $b$ is lower bound of $A$, as $b-\delta \lt b $ and $b-\delta \in A=(0,1)$.
So, $0$ is the greatest lower bound of $A$.
