$\newcommand{\Int}{\text{Int}}$ $\newcommand{\Bd}{\text{Bd}}$


Let $A=\{x\times y|0<x^2-y^2\leq1\}$. Find the boundary and the interior of $A$.

Attempt Solution

We claim that $\Int\left(E\right)=\left\{ x\times y|0<x^{2}-y^{2}<1\right\} $. To prove this, it suffices to prove that any point $\left(x,y\right)\in\left\{ x\times y|0<x^{2}-y^{2}<1\right\} $ is in the interior of $E$ and any point $\left(x,y\right)\in\left\{ x\times y|x^{2}-y^{2}=1\right\} $ is not in the interior of $E$. Suppose $x^{2}-y^{2}=a$ for some $a\in\left(0,1\right).$ Then by geometry (or computation), $\exists\varepsilon_{1},\varepsilon_{2}>0$ such that for $\forall\left(x,y\right)\in\left(x-\varepsilon_{1},x+\varepsilon_{2}\right)\times\left(y-\varepsilon_{2},y+\varepsilon_{2}\right):=U$, we have $x^{2}-y^{2}\in\left(0,1\right)$. Therefore, $U$ is open set containing $\left(x,y\right)$ and $U\subset E$ and as a result $\left(x,y\right)\in\Int\left(E\right).$ Now suppose $x^{2}-y^{2}=1$ and let $U$ be any open set that contains $\left(x,y\right).$ Then we can find a basic element $B=\left(a,b\right)\times\left(c,d\right)$ such that $\left(x,y\right)\in\left(a,b\right)\times\left(c,d\right)\subset U$. Then $\left(x-\varepsilon,x+\varepsilon\right)\times y\subset\left(a,b\right)\times\left(c,d\right)$ for some $\varepsilon>0$. Then $\left(x+\varepsilon/2\right)^{2}-y^{2}>1$ and thus $U\nsubseteq E$ and so $\left(x,y\right)\notin\Int\left(E\right).$ Therefore, $\Int\left(E\right)=\left\{ x\times y|x^{2}-y^{2}\in\left(0,1\right)\right\} .$

Similarly, we claim that $\bar{E}=\left\{ x\times y|0\leq x^{2}-y^{2}\leq1\right\} .$ We prove this by proving that $\left\{ x\times y|x^{2}=y^{2}\right\} \in E'$ and $\left\{ x\times y|x^{2}-y^{2}>1\mbox{ or }x^{2}-y^{2}<0\right\} \notin E'$. Let $U$ be any open set containing $\left(x,y\right)$ such that $x^{2}-y^{2}=0$ and then we can find a basic set $\left(a,b\right)\times\left(c,d\right)$ such that $\left(x,y\right)\in\left(a,b\right)\times\left(c,d\right)\subset U.$ Then $\exists\varepsilon>0$ such that $\left(x-\varepsilon,x+\varepsilon\right)\times y\subset\left(a,b\right)\times\left(c,d\right)\subset U$. As a a result, we can find some $x'\in\left(x-\varepsilon,x+\varepsilon\right)$ such that $x'^{2}-y^{2}\in(0,1]$. Therefore, $\big(U\backslash\left(x,y\right)\big)\cap E\neq\emptyset$, which implies that $\left(x,y\right)\in E'$. Now let $\left(x,y\right)\in\left\{ x\times y|x^{2}-y^{2}>1\right\} $. Then by similar reasoning as previous argument, $\exists\varepsilon_{1},\varepsilon_{2}>0$ such that for $\forall\left(x,y\right)\in\left(x-\varepsilon_{1},x+\varepsilon_{1}\right)\times\left(y-\varepsilon_{1},y+\varepsilon_{2}\right)$ we have $x^{2}-y^{2}>1,$ which implies $\left(x,y\right)\notin E'$. By exactly the same line, we can show that $\left(x,y\right)\notin E'$ for $\forall\left(x,y\right)\in\left\{ x\times y|x^{2}-y^{2}<0\right\} .$ As a result, we have $\bar{E}=E\cup E'=\left\{ x\times y|0\leq x^{2}-y^{2}\leq1\right\} .$ Hence, $\Bd\left(E\right)=\bar{E}\backslash\Int\left(E\right)=\left\{ x\times y|x^{2}-y^{2}=0\mbox{ or }x^{2}-y^{2}=1\right\} $.


I don't know how to argue more elegantly for this problem. Could sometime take a look at my proof and tell me if I did anything wrong? I really appreciate it! I think the intuition of this problem is quite easy, but it is not easy to formalize the argument..

  • $\begingroup$ Is $x\times y$ a new notation for a pair? $\endgroup$ – Nex Nov 3 '15 at 9:54
  • $\begingroup$ @Nex It's Munkres' notation. I'm not a fan, but one does see it on this forum... $\endgroup$ – Henno Brandsma Nov 3 '15 at 18:49
  • $\begingroup$ @HennoBrandsma I would be interested to know what motivated such an ambigous notation. If $S$ and $T$ are sets how do we decide whether $S\times T$ should be a pair or a set of pairs? $\endgroup$ – Nex Nov 3 '15 at 18:54
  • $\begingroup$ @Nex I think it denotes a pair if it's not capitalized and a set of pair if capitalized. $\endgroup$ – Mathemagician Nov 3 '15 at 19:51

To see openness of the interior, define $f(x,y) = x^2 - y^2$ (from the plane to the reals). One can argue from standard theorems that $f$, or any polynomial, is continuous. Then the set is just $f^{-1}[(0,1)]$, which is open by continuity.

Same for closedness and $f^{-1}[[0,1]]$.

One does have to argue a bit (like you did) that points with $x^2 - y^2 = 1$ are not in the interior (I agree with that argument), and that points with $x^2 - y^2 = 0$ are in the closure, as you did. So you can keep those smaller parts and leave the rest of the cases, or appeals to geometry and replavce them with the continuous function argument above. Because when the first set is open, and the points with $x^2 - y^2= 1$ are not in the interior, it is automatically the largest open subset, and likewise for the closure.


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