Proof that a subset of the plane is an open set. I don't have a specific book that I am learning from because I am trying to learn on my own but my used definition of an open set $A$ is that for every $a\in A$ there exists an $\varepsilon>0$ such that the open ball $B(a,\varepsilon)\subset A$
Let $\boldsymbol{A}=\{(x,y)\in \mathbb{R}^2|0<x<1,0<y<2\}\subset \mathbb{R}^2$. Prove that $\boldsymbol{A}$ is an open set. 
My attempt: Let $P=(a,b)\in \boldsymbol{A}$. We can choose a radius $\varepsilon<\min\{|a-1|,|b-2|,a,b\}$. (This choice because they are the distances from the lines that bound the set $\boldsymbol{A}$ and it certainly seems to work). Now I think I would have to prove that if I take a point $(x,y)\in B(P,\varepsilon)$, then $(x,y)\in \boldsymbol{A}$, but I don't see how I can do this. I have tried to say that $\sqrt{(a-x)^2+(b-y)^2}<\varepsilon$ and the tried to somehow show in cases that always $0<x<1$ and $0<y<2$ but it doesn't seem to lead anywhere. I also don't even know how much work is needed to finish this proof because I haven't really seen any examples done. 
Can someone help me finish this argument or give a more elegant one alternatively(should still be from the definition)?
 A: You're definitely on the right track. For each of the four inequalities you need to show, you need only to expand the inequality you have so that it ignores the irrelevant stuff. For example, to show $0<x$, 
$$a-x<\sqrt{(a-x)^2+(b-y)^2}<\varepsilon<a$$
A: Let $(x,y) \in B(P,\epsilon),$ then $$(a-x)^2+(b-y)^2<\epsilon^2.$$ In particular $$(a-x)^2<\epsilon^2 \text{ and } (b-y)^2<\epsilon^2.$$ This gives $$|a-x|<\epsilon<a,|a-1|.$$So $0<x<1.$ Similary $0<y<2$ and thus $(x,y)\in B(P,\epsilon).$
A: Suppose that $\langle x,y\rangle\in B(P,\epsilon)$. Then $|x-a|<\epsilon$, so $a-\epsilon<x<a+\epsilon$. You chose $\epsilon$ so that $a>\epsilon$, so $a-\epsilon>0$, and therefore $x>0$. You also chose it so that $1-a>\epsilon$, so $a+\epsilon<1$, and therefore $x<1$. Thus, $0<x<1$. You can apply the same kind of reasoning to show that $0<y<2$ and hence that $\langle x,y\rangle\in\boldsymbol{A}$.
In case it’s not clear that $\langle x,y\rangle\in B(P,\epsilon)$ implies that $|x-a|<\epsilon$, observe that
$$|x-a|=\sqrt{(x-a)^2}\le\sqrt{(x-a)^2+(y-b)^2}<\epsilon\;.$$
