(b) Give examples to show that the converse fails to hold if either of the requirements of X being compact or fn being monotone increasing is deleted.
Proof: (a) For each n, let Kn={x∈X | |fn(x)−f(x)|≥ε}. It is routine to show Kn is closed, by showing its complement is open: Suppose |fn(x)−f(x)|<ε. Then f(x)∈(fn(x)−ε,fn(x)+ε)=V, so let U be a neighborhood of f(x) contained in V, and we see f−1(U)⊆X is open such that f(U)⊆V so that U is a neighborhood of x also contained in the complement of Kn. Now, since (fn) is monotone increasing, and hence f(x)≥fn(x) by convergence, we see Kn+1⊆Kn for all n. If we assume each Kn is nonempty, then finite intersections ∩Ki=Kmax i are also nonempty, so since X is compact, there exists x∈X such that x∈Kn for all n, i.e. |fn(x)−f(x)|≥ε for all n. However, this is impossible since fn(x)→f(x) for all x. Hence KN=ø for some N, hence Kn=ø for all n≥N, that is, |fn(x)−f(x)|<ε for all x, for sufficiently large n.
(b) Suppose X need not be compact; then the sequence of functions (fn) defined by fn(x)={0:x≤nn−x:x>n is seen to be monotone increasing with fn(x)→0 for all x, yet does not converge uniformly to the zero function.
Suppose fn need not be monotone increasing; then let [0,1]=∪[an,bn] be any countably infinite partition of [0,1] into disjoint closed intervals where an≠bn for all n. Then when cn=an+bn2, define a sequence of functions (fn) by fn(x)={0:x∉[an,bn]x−ancn−an:x∈[an,cn]x−bncn−bn:x∈[cn,bn] It is clear by the pasting lemma that each fn is a continuous function [0,1]→[0,1]. As well, it is clear fn(x)≠0 for at most one value of n, so that fn(x)→0 for each x. However, since fn(cn)=1 for each n, we do not have uniform convergence to the zero function. ◻
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