312 CHAPTER 15. SEQUENCES, COMPACTNESS, CONTINUITY
yx
rr1
B(x,r)
yx
rr1
D(x,r)
The next theorem includes the main ideas for a set to be closed. It says that closed is toretain all limits of sequences which are contained in A.
Theorem 15.2.4 A nonempty set A is closed if and only if whenever xk ∈ A andlimk→∞xk = x, it follows that x ∈ A. In other words, the set is closed if and only if everyconvergent sequence of points of A converges to a point of A.
Proof: Suppose A is closed and suppose limk→∞xk = x. Does it follow that x ∈ A? Ifnot, then since A is closed, its complement is open and so there is a ball B(x,r) containedin AC. However, this contradicts the assertion that x is the limit of the sequence. Indeed,xk must be in B(x,r) for all k sufficiently large.
Conversely, suppose A retains all limits of convergent sequences. Is A closed? In otherwords, is its complement AC open? Suppose x ∈ AC. Is B(x,r) ⊆ AC for small enoughpositive r? If not, then B
(x, 1
k
)contains a point of A called xk for each k = 1,2, · · · . Thus
x is a limit of the sequence {xk} and so x ∈ A after all. Hence AC must indeed be openand so, by definition, A is closed.
15.3 Cartesian ProductsRecall R2 consists of ordered pairs (x,y) such that x ∈ R and y ∈ R. R2 is also written asR×R. In general, the following definition holds.
Definition 15.3.1 The Cartesian product of two sets A×B, means
{(a,b) : a ∈ A, b ∈ B} .
If you have n sets A1,A2, · · · ,An
n
∏i=1
Ai = {(x1,x2, · · · ,xn) : each xi ∈ Ai} .
Now suppose A ⊆ Rm and B ⊆ Rp. Then if (x,y) ∈ A× B, x= (x1, · · · ,xm), andy = (y1, · · · ,yp), the following identification will be made.
(x,y) = (x1, · · · ,xm,y1, · · · ,yp) ∈ Rp+m.
Similarly, starting with something in Rp+m, you can write it in the form (x,y) wherex ∈ Rm and y ∈ Rp. The following theorem has to do with the Cartesian product of twoclosed sets or two open sets. Also here is an important definition.
Definition 15.3.2 A set, A ⊆Rp is said to be bounded if there exist finite intervals,[ai,bi] such that A ⊆ ∏
pi=1 [ai,bi] .