Multivariable Advanced Calculus

92 CONTINUOUS FUNCTIONSlet C ∩ (−∞, x) ≡ A, and C ∩ (x, ∞) ≡ B. Then C = A ∪ B and the sets A and Bseparate C contrary to the assumption that C is connected.Conversely, let I be an interval. Suppose I is separated by A and B. Pick x ∈ Aand y ∈ B. Suppose without loss of generality that x < y. Now define the set,S ≡ {t ∈ [x, y] : [x, t] ⊆ A}and let l be the least upper bound of S. Then l ∈ A so l /∈ B which implies l ∈ A. Butif l /∈ B, then for some δ > 0,(l, l + δ) ∩ B = ∅contradicting the definition of l as an upper bound for S. Therefore, l ∈ B which impliesl /∈ A after all, a contradiction. It follows I must be connected. This yields a generalization of the intermediate value theorem from one variablecalculus.Corollary 5.3.9 Let E be a connected set in a normed vector space and supposef : E → R and that y ∈ (f (e 1 ) , f (e 2 )) where e i ∈ E. Then there exists e ∈ E such thatf (e) = y.Proof: From Theorem 5.3.5, f (E) is a connected subset of R. By Theorem 5.3.8f (E) must be an interval. In particular, it must contain y. This proves the corollary.The following theorem is a very useful description of the open sets in R.Theorem 5.3.10 Let U be an open set in R. Then there exist countably manydisjoint open sets {(a i , b i )} ∞ i=1 such that U = ∪∞ i=1 (a i, b i ) .Proof: Let p ∈ U and let z ∈ C p , the connected component determined by p. SinceU is open, there exists, δ > 0 such that (z − δ, z + δ) ⊆ U. It follows from Theorem5.3.4 that(z − δ, z + δ) ⊆ C p .This shows C p is open. By Theorem 5.3.8, this shows C p is an open interval, (a, b)where a, b ∈ [−∞, ∞] . There are therefore at most countably many of these connectedcomponents because each must contain a rational number and the rational numbers arecountable. Denote by {(a i , b i )} ∞ i=1the set of these connected components. This provesthe theorem. Definition 5.3.11 A set E in a normed vector space is arcwise connected if forany two points, p, q ∈ E, there exists a closed interval, [a, b] and a continuous function,γ : [a, b] → E such that γ (a) = p and γ (b) = q.An example of an arcwise connected topological space would be any subset of R nwhich is the continuous image of an interval. Arcwise connected is not the same asconnected. A well known example is the following.{(x, sin 1 ) }: x ∈ (0, 1] ∪ {(0, y) : y ∈ [−1, 1]} (5.2)xYou can verify that this set of points in the normed vector space R 2 is not arcwiseconnected but is connected.Lemma 5.3.12 In a normed vector space, B (z,r) is arcwise connected.