Let $A = (0,1]$. Then$\text{ inf}(A) = 0$ I posted before about this proposition and I thought I got it right but then I was told that it is still wrong so I am really confused again.. 
Here is my proof
Proof : Let $A = (0,1]$
Here, since $0 ≤ a$ for all $a ∈ A$, $0$ is an lower bound
Now, suppose that there is some $b ∈ R$ such that $b$ is a lower bound for $A$ and $b > 0$
Then assuming $b ≤ 1$, since $b ∈ A$, b can not be a lower bound of $A$.
Note that $$\frac b2 \in A \;\;\;and\;\;\;\frac b2<b=\text{inf}(A)$$which contradicts the property that $\text{inf}(A)$ is smaller than any element of $A$. 
Hence $0$ is an $\text{inf}(A)$.
Please help!

This is good. Alternatively, for a direct proof, note $A$ is non-empty and bounded below. Every lower bound $c$ of $A$ must satisfy $c \notin \mathbb{R_+}$ since for all $\epsilon \in \mathbb{R_+}$ there is some $n$ such that $\epsilon > \frac{1}{n}$ and $\frac{1}{n} \in A$.  Since $\max \mathbb{R} \setminus \mathbb{R_+} = 0$, and $0$ is a lower bound of $A$, it follows $\inf A = 0$.  
