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Can someone teach me how to find interior, exterior and boundary of these two sets in the plane, $\mathbb R^2$? The metric is $d_2 (x,y)=\sqrt{(x_1-y_1)^2 +(x_2-y_2)^2}$, where $x = (x_1, x_2)$ and $y = (y_1, y_2)$.

$A = \{(x,y): xy \neq 0\}$

$B = \{(x,y):x^2+y^2 <1 \text{ and } x,y \in \mathbb Q\}$

It's really confusing to me. Thanks for your help.

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I fixed some Latex Problems; please edit again if I changed anything fundamental to your problem. – William Jun 12 '12 at 6:41
thank you so much. It's all correct. I'm very new to this website so i don't know how to type all the math symbols. Thanks again. – Johnny Jun 12 '12 at 6:43

Since $xy \neq 0$ if and only if $x \neq 0$ and $y \neq 0$. Thus $A$ is $\mathbb{R}^2$ missing the $x$ and $y$ axis. $A$ is open. Hence its interior is $A$, itself. The exterior is the interior of the complement of $A$. The complement of $A$ is just the union of the $x$ and $y$ axis. It contains no open set, so it has empty interior. The exterior of $A$ is the $\emptyset$. The boundary is the set of all points such that every neighborhood intersects $A$ and its complement.The only point with these property is the union of the $x$ and $y$ axis.

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Thanks a lot I got it now! – Johnny Jun 12 '12 at 7:02
Your answer on $B$ is not correct. – T. Eskin Jun 12 '12 at 21:13
I forgot that $x,y \in \mathbb{Q}$. – William Jun 12 '12 at 21:16

Usually when sets have such expressions it is very useful to check if working with continuous functions is possible, as it is in the case of $A$.

Note that if $f:\mathbb{R}^{2}\to \mathbb{R}$ is given by $f(x,y)=xy$, then $f$ is continuous and $A=f^{-1}(\mathbb{R}\setminus\{0\})$. Hence $A$ is open as a preimage of an open set under a continuous function. So $A$ equals its own interior. Moreover, for any point in the plane with either $x=0$ or $y=0$ the open ball $B((x,y),r)$ contains points $(a,b)$ such that $ab\neq 0$. Hence the boundary of $A$ consists of the $x$-axis and the $y$-axis. This leaves the exterioir of $A$ to be empty.

Note that $B=B(\bar{0},1)\cap \mathbb{Q}^{2}$, where $B(\bar{0},1)$ is the open $1$-radius ball around origin $\bar{0}$. So $B$ is basicly the rational coordinate points of the open unit ball. The interior of $B$ is empty since it contains no open balls: every open ball in $\mathbb{R}^{2}$ contains points with irrational coordinates and with rational coordinates. This being said, every open ball around a point in $C:=\{(x,y):x^{2}+y^{2}\leq 1\}$ has a non-empty intersection with $B$. This implies that $C$ is a subset of $B$'s boundary. Since the interior $B$ is empty then the boundary of $B$ equals its closure $\mathrm{cl}(B)$. Now since $C$ is closed and $B\subset C\subset \mathrm{cl}(B)$ it follows that $C$ is in fact the whole closure (since the closure is the smallest closed set containing $B$), and thus the boundary of $B$ is $C$ as well. Finally, the exterior of $B$ is what is left in $\mathbb{R}^{2}$, i.e. the set $\{(x,y):x^{2}+y^{2}>1\}$.

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