- #1
alanlu
- 65
- 0
I'm trying to show that continuous f : [a, b] -> R implies f uniformly continuous.
f continuous if for all e > 0, x in [a, b], there exists d > 0 such that for all y in [a, b], ¦x - y¦ < d implies ¦f(x) - f(y)¦ < e.
f uniformly continuous if for all e > 0, there exists d > 0 such that for all x and y in [a, b], ¦x - y¦ < d implies ¦f(x) - f(y)¦ < e.
I constructed
A(d) = { u in [a, b] : x, y in [a, u], ¦x - y¦ < d implies ¦f(x) - f(y)¦ < e }
A = Ud > 0 A(d)
And I think I need to show sup A = b and b is in A, but I'm stuck.
f continuous if for all e > 0, x in [a, b], there exists d > 0 such that for all y in [a, b], ¦x - y¦ < d implies ¦f(x) - f(y)¦ < e.
f uniformly continuous if for all e > 0, there exists d > 0 such that for all x and y in [a, b], ¦x - y¦ < d implies ¦f(x) - f(y)¦ < e.
I constructed
A(d) = { u in [a, b] : x, y in [a, u], ¦x - y¦ < d implies ¦f(x) - f(y)¦ < e }
A = Ud > 0 A(d)
And I think I need to show sup A = b and b is in A, but I'm stuck.