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 $A$.


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 $$ -x_1 < y_1 - x_1 < x_1 \quad \Rightarrow \quad 0 < y_1. $$

  • $\begingroup$ Thanks for the help, Dylan. For (a), is there a more formal way to show $B_r(x)\subset A$ where $r=min(x_1,x_2)$? Graphically, it makes sense -- a ball of such radius $r$ around any point $(x_1,x_2)$ where $x_1,x_2>0$ can't possibly lie outside the first quadrant. I'm just not sure how to formalize that. $\endgroup$
    – Emir
    Jan 30 '12 at 7:40
  • $\begingroup$ @Emir You can certainly formalize it--I didn't mean to imply that pictures are the same as proofs! I just wanted to indicate the intuition. It shouldn't be so bad, but unfortunately I must sleep now--I will say something in the morning if no one else has chimed in by then. $\endgroup$ Jan 30 '12 at 7:45
  • $\begingroup$ @Emir I've added a quick exercise that should make this clearer. Let me know if it causes trouble. $\endgroup$ Jan 30 '12 at 15:45
  • $\begingroup$ For (1), you note that $d(x,y)=\sqrt{(y_{1}-x_{1})^{2}+(y_{2}-x_{2})^{2}}$, thus $d(x,y)\geq|y_{1}-x_{1}|+|y_{2}-x_{2}|$. I am confused as to why you have $x_1\geq r$ in (2) though? $\endgroup$
    – Emir
    Jan 30 '12 at 19:45
  • $\begingroup$ @Emir We chose $r$ to be the lesser of $x_1$ and $x_2$. $\endgroup$ Jan 30 '12 at 22:13

For (a), here's two different ways of showing that the set is open:

$\#1$: If $pr_{1}:\mathbb{R}^{2}\rightarrow \mathbb{R}$ and $pr_{2}:\mathbb{R}^{2}\rightarrow \mathbb{R}$ are projections to the first and second component respectively, then they are continuous functions and:

$$A:=\{(x,y)\in \mathbb{R}^{2}:x>0,y>0\}=pr_{1}^{-1}(]0,\infty[)\cap pr_{2}^{-1}(]0,\infty[)$$

is open as the intersection of two open sets. (the preimage of an open set under continuous function is open).

$\#2$: Using the definition, if $(x,y)\in A$, then $x>0$ and $y>0$. Denote $r=\min\{x,y\}$. Now $B_{r}(x,y)\subset A$, because if $(x',y')\in B_{r}(x,y)$ then $|x-x'|<r\leq x$ and $|y-y'|<r\leq y$, so it can't be that $x'\leq 0$ or $y'\leq 0$.

For (b), recall the discrete metric: $d(x,y)=1$ if $x\neq y$ and $d(x,y)=0$ if $x=y$. If $(X,d)$ is a space with discrete metric and $A\subset X$, then for any $a\in A$ we have $B_{\frac{1}{2}}(a)=\{a\}\subset A$. So $A$ is an open set.


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