Prove that the following are open sets:
(a) the “first quadrant,” $\{(x,y)\in\mathbb{R^2}\mid x>0 \text{ and }y>0\}$
(b) any subset of a discrete metric space
I'm using the following definition of an open set in a metric space $(X,d)$: a set $A$ is open if $\forall a\in A,\exists r>0$ s.t. $B_r(a)\subseteq A$.
For (a), the metric space is $(\mathbb{R^2},d)$. For $r\in\mathbb{R}$ and $r>0$ An open ball of radius $r$ around $x_o\in\mathbb{R^2}$ is $B_r(x_o)=\{x\in\mathbb{R^2}\mid d(x,x_o)
Attempt at (a):
Let $x_o\in\{(x,y)\in\mathbb{R^2}\mid x>0\text{ and }y>0\}$. Suppose that $\forall r\in\mathbb{R}$, $B_r(x_o)\not\subseteq A$. Then, $\exists r\in\mathbb{R}$ s.t. $B_r(x_o)\subseteq A^{c}$. But that means $x_o$ has negative components even though its in the first quadrant?
Attempt at (b):
For (b), Let $a_o\in A\subseteq X$. Then, $d(a_o,a)<2\in\mathbb{R},\forall a\in A$. Since $a_o$ was arbitrary there exists an $r>0$ for all $a\in A$ such that the open ball $B_r(a)$ is a subset $A$.