Let $A = (0,1]$. Then $\inf(A) = 0$ I am having a problem with this statement. I am trying to prove that 0 is the greatest lower bound by showing that every lower bound greater than 0 is a contradiction but I can't figure our how.
Proposition.
Let $A = (0,1]$. Then $\inf(A) = 0$
Proof.
Here, $0 < a$ for all $$a \in A$$
So $0$ is a lower bound for $A$.
Now suppose there is some $$b \in R$$ such that b is a lower bound for $A$ and $b>0$
Then, assuming $b≤1$, since $$b \in A$$ b can not be a lower bound (is this part wrong? if right, why?)
Hence $0$ is the greatest lower bound and $\inf(A)$.

You're right. If $b \in A$ then $b$ cannnot be a lower bound, since, for exemple, $\frac{b}{2} \in A$ and $\frac{b}{2} < b$.
