Let $C[0,1]$ denote the metric space of all continuous functions $f:[0,1] \rightarrow \mathbb{R}$, with the metric
$d(f,g)=\sup |f(x) – g(x)|$ for $0\leq x \leq 1$
Show that $C[0,1]$ is a complete metric space
So far I have started with this:
Since $f$ is given to be continuous then it follows that for every sequence $(x_n)$ in $[0,1]$ which converges to every $x \in [0,1]$, then the sequence $(f(x_n))$in $\mathbb{R}$ converges to $f(x)$ for every $x\in [0,1]$ thus the sequence $(f(x_n))$ is Cauchy for every sequence $(x_n)$ in $[0,1]$ which converge to all points $x\in [0,1]$. Overall, by the definition of continuity the sequence $(f(x_n))$ is Cauchy for every $(x_n)\in [0,1]$ that converges to every $x\in[0,1]$.
And since every $(f(x_n))$ is Cauchy then $f$ is complete in $C$
I'm just wondering what I am doing wrong and what I can do to steer the proof in the right direction.
Thanks!
Best Answer
Judging from your proof, you've completely misunderstood the problem. You're supposed to show that every Cauchy sequence $(f_n)_{n = 1}^\infty$ in $C[0,1]$ converges to an element $f$ of $C[0,1]$. Given a Cauchy sequence $(f_n)$ in $C[0,1]$, we have for each $x\in [0,1]$, $(f_n(x))_{n = 1}^\infty$ is Cauchy in $\Bbb R$. Since $\Bbb R$ is complete, there exists a function $f : [0,1] \to \Bbb R$ such that $\lim\limits_{n\to \infty} f_n(x) = f(x)$ for all $x\in [0,1]$. Now you need to show that $f \in C[0,1]$.