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!