0
$\begingroup$

Suppose $\bigl\{(x,y)\mid x^2+y^2<1\bigr\}$ is a subset of $\Bbb R\times\Bbb R$, where $\Bbb R$ is the set of real numbers.

Is the given set also the cartesian product of two subsets of $\Bbb R$?

Does

$$X=\{x\in\Bbb R\mid 1>x>-1\}\\Y=\left\{y\in\Bbb R\mid\left((\forall x\in X)\left[-\sqrt{1-x^2}<y<\sqrt{1-x^2}\right]\right)\right\}$$

define infinitely many pairs $(x,y)$ for given $x$, such that $y$ suffices the condition in $Y$?

Also, could there be two other sets (except of just, in this case, replacing "$x$" with "$y$", "$y$" with "$x$", "$X$" with "$Y$" and "$Y$" with "$X$"), whose cartesian product yields the given subset?

$\endgroup$
  • $\begingroup$ Very Clever. Copied the images. How much time you put in doing it? $\endgroup$ – MonK Aug 22 '14 at 14:35
  • $\begingroup$ @MonK I can't find those images on other sites by using google search-by-image. And they probably won't be taken from math.stackexchange since they're not rendered in LaTeX. So I'm not certain that they've been copy-pasted. $\endgroup$ – Jam Aug 22 '14 at 14:37
  • $\begingroup$ No, I think you also din't get it. He has an image on his system. He took snip images out of it, put it over here in places in bits and pieces :) $\endgroup$ – MonK Aug 22 '14 at 14:39
  • $\begingroup$ Welcome to Math.SE! What are your thoughts on the problem so far? What have you tried? The more we know about your efforts and thinking, the easier it will be for us to help you (and the more likely it will be that people will want to help you). $\endgroup$ – Cameron Buie Aug 22 '14 at 14:43
  • $\begingroup$ Why do you think I copied them? I used Word, Snapping Tool (in windows 7) and microsoft Paint to save the pics. Anyhow, this problem is given in the first set of exercises about cartesian products, so really I've no experience about it, Cameron Buie. $\endgroup$ – availanche Aug 22 '14 at 14:43
4
$\begingroup$

If you define $Y$ that way, then $Y=\{0\}$. That's because when $y\neq 0$, with $-1<y<1$, then for $x=\sqrt{1-y^2}$, $x\in X$ it is obviously not true that $-\sqrt{1-x^2}<y<\sqrt{1-x^2}$. So $y\notin Y$.

If $U=X\times Y$ then if $(x_1,y_1),(x_1,y_2),(x_2,y_1)\in U$ then $(x_2,y_2)\in U$. When $X,Y\subseteq \mathbb R$, this can be read as:

If three corners of an "axis-aligned" rectangle are in $U$, then the fourth corner is, too.

Now, $U=\{(x,y)\mid x^2+y^2<1\}$ is a unit disk. Can you think of an axis-aligned rectangle with three points inside $U$ but the fourth outside?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @availanche Nope, there is such a rectangle. Try again. $\endgroup$ – Thomas Andrews Aug 22 '14 at 15:47
  • $\begingroup$ Such a rectangle does not exist, because the fourth vertex shares coordinates of the already existing 3 vertices, which are, at any order, define a point that is not outside the unit disk. I wanted to form ordered pairs, where, if the first coordinate is x, the second coordinate is limited by -sqrt(1-x^2) and sqrt(1-x^2). Does the cartesian product of two closed intervals always geometrically correspond to a rectangle? $\endgroup$ – availanche Aug 22 '14 at 16:14
  • $\begingroup$ Sorry, you are still wrong. Draw a circle. Can you create a rectangle with three points inside the circle and the fourth outside? $\endgroup$ – Thomas Andrews Aug 22 '14 at 16:26
  • $\begingroup$ Well, yes, I succeeded in finding infinitely many such surprising rectangles, in regard to the unit circle: take the points (0.5, -0.5), (0.5, -0.7), (g, -0.5) and (g, -0.7), where -(sqrt (51))/10>g>-(sqrt(3))/2...Oh I see, in regards to "If three corners of an "axis-aligned" rectangle are in U, then the fourth corner is, too." theorem, it means that a circle cannot be the result of any cartesian product, hence disks cannot be so as well. $\endgroup$ – availanche Aug 22 '14 at 17:17
  • $\begingroup$ You can just use $(0,0)$, $(x,0)$, $(0,y)$ and $(x,y)$ with $x,y\in(-1,1)$ and $x^2+y^2>1$.... $\endgroup$ – Thomas Andrews Aug 22 '14 at 17:21
0
$\begingroup$

The given set is not cartesian product of two subsets X,Y of the reals;

Supoosing X,Y are subsets of the reals whose Cartesian product is the unit disk (excluding the unit circle), it also means that their cartesian product corresponds to an axis-aligned rectangle.

In this case, the following theorem holds: "If three corners of an "axis-aligned" rectangle are in U, then the fourth corner is, too" (worded by Thomas Andrews), where U=XxY.

But, in case of circle with radius r, one may choose 3 points, each of which is a subset of the cartesian product of X and Y: (0,0), (x,0), (0,y) and (x,y), where x,y belong to the opened interval (-1,1), and x^2+y^2>r^2 (specifically, it would mean that at least one of |x|, |y|, must be greater than r/sqrt(2), otherwise their sum cannot be greater than r^2).

Therefore, we deduce that the given disk (x^2+y^2<1) cannot be the cartesian product of two subsets of R.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It would be better and clear if you have used images in your answer too :) $\endgroup$ – Freddy Aug 22 '14 at 18:08
  • $\begingroup$ Here is some relevant demonstration from desmos (finding such configuration for the unit circle itself: desmos.com/calculator/yglrzwlmkt $\endgroup$ – availanche Aug 22 '14 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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