In addition, we say that the domain of a function is bounded if there is a number R > 0 such that the domain is inside of the circle centered at ( 0,0) with radius R. For example, the domain in example 5 is bounded because it is itself a circle centered at the origin.
Conversely, a set is unbounded if it cannot be contained in any circle centered at the origin. For example, set {(x,y) | y > 2x } is unbounded since no circle centered at the origin can contain all the points (x,y) for which y > 2x.
EXAMPLE 6 Determine if the domain of f( x,y) = ln( xy+1) is bounded or unbounded.
Solution: To begin with, we must have xy+1 > 0, which requires that xy > -1. If x < 0, then xy > -1 is the same as y < -1/x. If x ³ 0, then xy > -1 is the same as y > -1/x. We can combine the two cases by using the symbol for a union of sets, which is È. That is,
The domain of the function is shown below:
dom(f) = ì
í
î( x,y) | x < 0 and y <
-1
x ü
ý
þÈ ì
í
î( x,y) | x ³ 0 and y >
-1
x ü
ý
þ
Clearly, it cannot be contained in any circle centered at the origin. Therefore, dom( f) is unbounded.
In addition, we say that a set R is connected if any two points in R can be joined by a curve which is contained in R:
EXAMPLE 6 Determine if the domain of the following function is connected.
f(x,y) = y
x2-1
Solution: The domain of f is the set of points (x,y) for which x2-1 ³ 0, which is where x2 ³ 1. This implies that either x £ -1 or x ³ 1, so that the domain is the two strips shown below:
Clearly, the domain is not connected.
