I am aware of this question but I want to request feedback on my proofs specifically. I am required to prove the following two conditionals:
- If $f:(0,1) \rightarrow R$ is uniformly continuous and $(x_{n})$ is a Cauchy sequence, then $f(x_{n})$ is a Cauchy sequence.
Proof: Suppose $(x_{n})$ is a Cauchy sequence. Let $\epsilon > 0$. Then, by definition, there exists $N \in N$ s.t. for all $m, n \geq N$, $| x_{n}-x_{m}| < \epsilon$. Now, pick $\delta = \epsilon$. Then, suppose $f$ is uniformly continuous on $(0,1)$. Then, $| x_{n}-x_{m}| < \delta$ implies $|f(x_{n}) – f(x_{m})| < \epsilon$ which shows that $f(x_{n})$ is a Cauchy sequence. I am a little apprehensive about the "which shows that $f(x_{n})$ is a Cauchy sequence." Is this proof accurate?
Secondly, I am required to prove:
- If $f:[0,\infty) \rightarrow R$ is continuous and $(x_{n})$ is a Cauchy sequence, then $f(x_{n})$ is a Cauchy sequence.
Proof: Suppose $(x_{n})$ is a Cauchy sequence. Then, $(x_{n})$ converges, that is, $\lim (x_{n}) = x$ exists. Note that $[0,\infty)$ is closed since its complement is open. So, $x \in [0,\infty)$. Suppose$f:[0,\infty) \rightarrow R$ is continuous. Then, $f(x_{n})$ must be convergent. Since all convergent sequences are Cauchy, $f(x_{n})$ must be Cauchy. Am I missing something in this proof? Thank you!
Best Answer
In your first proof, you use $\epsilon$ twice, but they are not the same epsilon.
I think I would begin with the $f(x)$ is uniformly continuous.
$\forall\epsilon >0, \exists \delta > 0,\forall x,y \in (0,1): |x-y|<\delta \implies |f(x)-f(y)| < \epsilon$
This establishes our epsilon. Now, we can work to keep the differences less than epsilon.
$\{x_n\}$ is Cauchy tells us there exists an $N >0$ such that $n,m>N \implies |x_n-x_m| < \delta$
Note that I have used delta instead of the traditional epsilon. This is the same delta from the line above.
Therefore $n,m>N \implies |f(x_n) - f(x_m)| < \epsilon.$
This is the same epsilon, and the proof flows more directly from givens to the conclusion.
The second proof still has some holes.
You state that $[0,\infty)$ is closed, but not why that is relevant. You state that $f(x_n)$ is convergent, without any justification. Invoke the definition of limits to show convergence.
$\forall\epsilon >0, |x_n-x|<\delta \implies |f(x_n) - f(x)|<\epsilon$