- #1
jmjlt88
- 96
- 0
Let X be a set, and let fn : X---> R be a sequence of functions. Let ρ be the uniform metric on the space RX. Show that the sequence (fn) converges uniformly to the function f:X--> R if and only if the sequence (fn) converges to f as elements of the metric space (RX, ρ). [Note: the ρ's should have a bar over them.]
I have a question concerning one direction of our implication.
[My attempt at the solution.]
Since I do not know how to make a d with a bar over it, I'll use ∂ to denote the standard bounded metric relative to d.
Proof:
Assume the sequence (fn) converges to f as elements of the metric space (RX, ρ).
Then, for any ε>0, there is a positive integer N such that ρ(fn,f)<ε for all n≥N.
That is, for n ≥ N, the sup{∂(fn(x),f(x))| x in X} < ε.
If ε≤1, our metrics are the same.
Hence for ε≤1, there exisits an integer N such that for n ≥ N, the sup{∂(fn(x),f(x))| x in X} < ε.
But since ∂(fn(x),f(x)) = d(fn(x),f(x)) for n ≥ N, d(fn(x),f(x)) ≤ sup{∂(fn(x),f(x))| x in X} < ε for all n≥N and x in X, which satisfies our condition for uniform convergence.
My question is about ε. Is this the right approach?
I have a question concerning one direction of our implication.
[My attempt at the solution.]
Since I do not know how to make a d with a bar over it, I'll use ∂ to denote the standard bounded metric relative to d.
Proof:
Assume the sequence (fn) converges to f as elements of the metric space (RX, ρ).
Then, for any ε>0, there is a positive integer N such that ρ(fn,f)<ε for all n≥N.
That is, for n ≥ N, the sup{∂(fn(x),f(x))| x in X} < ε.
If ε≤1, our metrics are the same.
Hence for ε≤1, there exisits an integer N such that for n ≥ N, the sup{∂(fn(x),f(x))| x in X} < ε.
But since ∂(fn(x),f(x)) = d(fn(x),f(x)) for n ≥ N, d(fn(x),f(x)) ≤ sup{∂(fn(x),f(x))| x in X} < ε for all n≥N and x in X, which satisfies our condition for uniform convergence.
My question is about ε. Is this the right approach?