# Math Help - The boundary of a two variable function's Domain

1. ## The boundary of a two variable function's Domain

Why the boundary of a two variable function's domain like $f(x,y) = \frac{y}{x^2}$
is $x=0$ and for $f(x,y) = x.y$ there's no boundary? According to the definition,
a point in a region $R$ is a boundary point if it is the center of every disk that
contains points that lie outside of $R$ as well as points that lie in $R$. My question for
the above problem is how the boundary is found? The book is vague about the
procedure for finding the boundary. It would be nice if anyone could explain it a bit.

2. Originally Posted by x3bnm
Why the boundary of a two variable function's domain like $f(x,y) = \frac{y}{x^2}$
is $x=0$ and for $f(x,y) = x.y$ there's no boundary? According to the definition,
a point in a region $R$ is a boundary point if it is the center of every disk that
contains points that lie outside of $R$ as well as points that lie in $R$. My question for
the above problem is how the boundary is found? The book is vague about the
procedure for finding the boundary. It would be nice if anyone could explain it a bit.
We cannot divide by 0, this simply does not exist.

Thus,

Anything of the form,

$g(x,y) = \frac{ h(x,y) } {p(x,y)}$

Has the restriction such that $p(x,y)$ cannot be equal to o.

When we have,

$f(x,y) = xy$ there is no diviser and as a result, this function is defined everywhere!

3. So are you saying that a function where it is undefined is the boundary of that function?

4. Originally Posted by x3bnm
So are you saying that a function where it is undefined is the boundary of that function?
I'm saying for a rational fraction the divisor cannot be equal to 0 and any solution that makes this equal to 0 is a vertical asymtote.

5. Originally Posted by x3bnm
So are you saying that a function where it is undefined is the boundary of that function?
No, he is saying nothing of the sort- and it is not the boundary of the function, it is the boundary of the domain of the function.

The "default" domain of a function, given only by a formula, is the set of all points such that we can calculate the formula. In the case of $f(x)= \frac{y}{x^2}$, the domain is "all (x,y) except for x= 0. That is, the domain is all of $R^2$ except the y-axis. Since that is a one-dimensional line, any small disk around a point on the y-axis contains some numbers that are not on the y-axis. That is, it contains some point on the y-axis, which are NOT in the domain, and some point not on the y-axis which ARE in the domain. That is why the y-axis is is the boundary of the domain.

The function f(x,y)= xy is defined for all points in the plane- its domain is all of $R^2$ and that is why its domain does not have a boundary.

Consider also the function $f(x,y)= \sqrt{x- 1}$. Since the square root of a negative number is not a real number, the domain of this function is the set of all points (x,y) with $x\ge 1$.

If x> 1, we can take a disk about any (x, y) with radius less than x- 1 and it will lie completely within the domain- it will contain only points in the domain so it is NOT a boundary point of the domain.

If x< 1, we can take a disk about any (x, y) with radius less than 1- x and it will lie completely in x< 1- no points in the disk will be in the domain so it is NOT a boundary point of the domain.

But if x= 1, any disk about (x, y) will contain points with x> 1, in the domain, and points with x< 1, not in the domain. That is, the boundary of the domain for this function is the line x= 1. Notice that that line is in the domain.

To get really complicated, I can define f(x,y) to be 1 if x and y are both rational numbers and not define it at all for (x,y) such that either x or y is irrational. Their the domain consists simply of (x,y) with both x and y rational. Because, between any two rational numbers there exist an irrational number and vice-versa, all points (x,y), whether in the domain or not, are boundary points of the domain!

6. HallsofIvy said:
No, he is saying nothing of the sort- and it is not the boundary of the function, it is the boundary of the domain of the function.

The "default" domain of a function, given only by a formula, is the set of all points such that we can calculate the formula. In the case of , the domain is "all (x,y) except for x= 0. That is, the domain is all of except the y-axis. Since that is a one-dimensional line, any small disk around a point on the y-axis contains some numbers that are not on the y-axis. That is, it contains some point on the y-axis, which are NOT in the domain, and some point not on the y-axis which ARE in the domain. That is why the y-axis is is the boundary of the domain.

The function f(x,y)= xy is defined for all points in the plane- its domain is all of and that is why its domain does not have a boundary.

Consider also the function . Since the square root of a negative number is not a real number, the domain of this function is the set of all points (x,y) with .

If x> 1, we can take a disk about any (x, y) with radius less than x- 1 and it will lie completely within the domain- it will contain only points in the domain so it is NOT a boundary point of the domain.

If x< 1, we can take a disk about any (x, y) with radius less than 1- x and it will lie completely in x< 1- no points in the disk will be in the domain so it is NOT a boundary point of the domain.

But if x= 1, any disk about (x, y) will contain points with x> 1, in the domain, and points with x< 1, not in the domain. That is, the boundary of the domain for this function is the line x= 1. Notice that that line is in the domain.

To get really complicated, I can define f(x,y) to be 1 if x and y are both rational numbers and not define it at all for (x,y) such that either x or y is irrational. Their the domain consists simply of (x,y) with both x and y rational. Because, between any two rational numbers there exist an irrational number and vice-versa, all points (x,y), whether in the domain or not, are boundary points of the domain!
HallsofIvy, I've a question about what you stated. Why did you chose radius of
disk to be less than $x - 1$ for $x > 1$ and $1 - x$ for $x < 1$ in your example and for
$f(x)= \frac{y}{x^2}$ radius to be any arbitrary length? Why is that? How did you figure out
the radius? Except this i understand everything you mentioned. Thanks for taking
the time in explaining that. I really appreciate it.