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 $B_r(x_o)=\{x\in\mathbb{R^2}\mid d(x,x_o)<r\}$ $\subseteq\mathbb{R^2}$. So basically I need to show that I can find an open ball with radius $r$ around any point in the first quadrant that is a subset of the first quadrant.
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 $A$.
Best Answer
For (a), I don't see how you argue that having each $B_r(x_0) \not\subset A$ implies that one of these is contained in the complement of $A$. It might be better to forget about contradiction: if $x = (x_1, x_2)$ is a point in the first quadrant, then let $r = \min(x_1, x_2)$ and show that $B_r(x) \subset A$. [Draw a picture!]
For (b), you'll have to say what your definitions are. When you say that a space is "discrete", that should mean that every subset is open. Perhaps you are starting from the standard discrete metric $$ d(x, y) = \begin{cases}1, & x \neq y \\ 0, &x = y\end{cases}; $$ in that case, given $A \subset X$ you should be able to show that for each $x \in A$ we have $B_1(x) = \{x\}$, and this is certainly a subset of $A$.
Added later. Let me replace your questions with more questions.
(1) Show that for general $x = (x_1, x_2)$ and $y = (y_1, y_2)$ in $\mathbf R^2$ we have $$ |y_1 - x_1|,\, |y_2 - x_2|\leq d(x, y). $$ (2) Thus, in the setting of (a), for $y \in B_r(x)$ we have for example $$ |y_1 - x_1| \leq d(x, y) < r \leq x_1 $$ and hence $$ -x_1 < y_1 - x_1 < x_1 \quad \Rightarrow \quad 0 < y_1. $$