- #1
rumjum
- 40
- 0
Homework Statement
This is not a homework question. I am solving this from the lecture notes that one of my friend's has got from last year.
If C(X) denotes a set of continuous bounded functions with domain X, then if X= [0,1] and fn(x) = x^n. Does the sequence of functions {fn} closed , bounded and compact? If not then why.
Homework Equations
In a compact metric space , every sequence has a convergent subsequence.
The Attempt at a Solution
My confusion is the following.
For x that belongs to [0,1), the sequence fn(x) -> - as n-> infinity. And
for x=1 , fn(x) ->1 as n -> infinity.
In this case, obviously fn(x) is bounded as for all n and all x, |fn(x)| <= 1. It is also closed. But, in the first place , how come is fn(x) continuous at x=1. Because for any n, for x->1, fn(x) ->1 from 0. Hence, we can choose an epsilon = 1/2 such that for delta = 0.5, we have |fn(x) - fn(1)| > epsilon.
Hence, fn(x) indicates a set of functions that is continuous only at X= [0,1) and not in [0,1]. Then, how come is fn(x) = x^n belong to C(X) , where X = [0,1].
Please explain.