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
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!