※ 引述《douglas0741 (23...........)》之銘言：
: Let {Fn} be a sequence of continuous on the interval [0,1]
: which converges uniformly on the rationals.
: show that there exists a continuous function , f on [0,1]
: such that {Fn} converges uniformly to f on [0,1]
: 希望有高手可以幫忙 謝謝
這樣可以嗎？
Let {F_n} be a sequence of continuous functions defined on [0,1] such that it
converges uniformly on Q∩[0,1]. Show that {F_n} converges uniformly on [0,1].
Proof. Given ε＞0, we want to find a positive integer N such that as m,n ≧N,
we have |F_n(x)－F_m(x)|＜ε for all x in [0,1]. By the assumption, we
know that for this ε, there is a positive integer N such that as m,n
≧ N, we have
|F_n(q)－F_m(q)|＜ε/6 for all q in Q∩[0,1]. (1)
Fixed y in [0,1]\Q and for each n, there exists a sequece of rational
numbers q_n (→y) such that
|F_n(q_n)－F_n(y)|＜ε/6 (2)
by continuity of F_n.
Moreover, since F_N(x) is uniformly continuous on [0,1], there exists
a δ＞0 such that as |x-x'|＜δ where x,x' in [0,1], we have
|F_N(x)－F_N(x')|＜ε/6 (3)
Hence as m,n≧N, choose q_n and q_m (in Q∩[0,1]) with |q_n－q_m|＜δ
so that (2) holds, and thus
|F_n(y)－F_m(y)| ≦ |F_n(y)－F_n(q_n)| ＜ ε/6 by (2)
＋ |F_n(q_n)－F_m(q_n)| ＋ ε/6 by (1)
＋ |F_m(q_n)－F_N(q_n)| ＋ ε/6 by (1)
＋ |F_N(q_n)－F_N(q_m)| ＋ ε/6 by (3)
＋ |F_N(q_m)－F_m(q_m)| ＋ ε/6 by (1)
＋ |F_m(q_m)－F_m(y)| ＋ ε/6 by (2)
So, |F_n(y)－F_m(y)|＜ε.
From above sayings, we complete it.□
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