Prove that the metric space $C[a,b]$ is complete.
Where $C[a,b]$ is the collection of continuous $f:[a,b] → R$ and $||f|| = sup_{x \in [a,b]} |f(x)|$, such that $\rho (f,g) = ||f - g||$ is a metric on $C[a,b]$.
attempt in proof:
Recall that a metric space X is said to be complete if and only if every Cauchy sequence $x_n \in X$ converges to some point in $X$.
Let $f_n$ be a Cauchy sequence in $ C[a,b]$, then $\forall \epsilon > 0,$ there is $N$ such that $||f_n - f_m|| < \epsilon$ for $n , m \geq N$ implies $||f_n - f_m || = sup |f_n - f_m | < \epsilon$.
Now for
$x \in [a,b]$, $|f_n - f_m | \leq sup_{x \in [a,b]} |f_n(x) - f_m(x)| < \epsilon$ for $n \geq N$.
Thus $f_n(x)$ converges uniformly to $f(x)$ .
And each $f_n$ is continous on $[a,b]$, and $f_n → f$ uniformly on $[a,b]$.
Thus, $f \in C[a,b]$. So $C[a,b]$ is complete.
Can someone please give some feedback? I don't know if I can conclude that $f_n$ converges uniformly to $f$. Can someone please help? Thank you in advance.
I don't think your argument works. There are two key ideas in this type of argument. One is to begin by finding the limit, and then prove it has whatever property we want. The other is the "$\varepsilon/3$" trick, which helps you relate regularity of the limit back to regularity of the members of the sequence.
In this case, the first idea goes through smoothly. You define $f$ to be the pointwise limit of the $f_n$. This exists because each sequence of numbers $f_n(x)$ is Cauchy (specifically, $|f_n(x)-f_m(x)| \leq \| f_n - f_m \|_\infty$), and the real numbers are complete. Now we use the second idea, by noting that for any $n$ we have
$$|f(x)-f(y)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|.$$
Now how can you control these three terms?