Why the boundary of a two variable function's domain like
is and for there's no boundary? According to the definition,
a point in a region is a boundary point if it is the center of every disk that
contains points that lie outside of as well as points that lie in . 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.
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 ofHallsofIvy 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!
disk to be less than for and for in your example and for
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.