Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
I fixed some Latex Problems; please edit again if I changed anything fundamental to your problem. –  William Jun 12 '12 at 6:41
1  
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
add comment

2 Answers

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 $\{(x,y):x^{2}+y^{2}>1\}$.

share|improve this answer
add comment

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.