Is $C(\mathbb R)$ Complete? I'm trying to prove an exercise from Carthers' book chapter10 of Real Analysis, problem claimed as,

where $C(\mathbb R)$ denote the infinity norm space of all continuous functions on real line.
I tried to use the hint. However, I got an counterexample that probably works for incomplete $C([-n,n])$. That is: $f_n(x) = x^{2n}$ does not converge to a continuous function in $C([-1,1])$.
Where is my fault?
 A: Let $(f_n)$ be a Cauchy sequence in $C(\Bbb R)$, and define $\Phi : [0, \infty) \to \Bbb R$ by the equation $\Phi(u) = \frac{u}{1+u}$. Given $\epsilon > 0$, there exists a positive integer $N$ such that $\|f_n - f_m\| < \epsilon$ for all $n, m \ge N$. Thus for every $k$, $\Phi(\|f_n - f_m\|_{C[-k,k]}) < \epsilon$ for all $n, m > N$. Since $\Phi$ is strictly increasing, then for each $k$, $\|f_n - f_m\|_{C[-k,k]} < \epsilon$ for all $n, m \ge N$. Therefore $(f_n)$ is Cauchy in $C[-k,k]$ for all $k$. Since $C[-k,k]$ is complete, there is a $g_k \in C[-k,k]$ such that $f_m \to g_k$ uniformly on $C[-k,k]$. The $g_k$ define a continuous function $g\in C(\Bbb R)$.
Choose a positive integer $n_0$ such that $\sum_{k=n_0+1}^\infty \frac{1}{2^k} < \epsilon$. The function $\Phi$ is bounded by $1$ so that $$\|f_n - g\| \le \sum_{k = 1}^{n_0} 2^{-k}\|f_n - g\|_{C[-k,k]} + \sum_{k = n_0 + 1}^\infty \frac{1}{2^k} \le \sum_{k = 1}^{n_0} 2^{-k}\|f_n - g_k\| + \epsilon$$ Since $\|f_n - g_k\|_{C[-k,k]} \to 0$ as $n \to \infty$ for every $k$, it follows that $$\limsup_{n \to \infty} \|f_n - g\| \le \epsilon$$ As $\epsilon$ is arbitrary, $f_n \to g$ in $C(\Bbb R)$.
A: I believe Carothers wanted the reader to use the metric for $C(\mathbb{R})$ which was defined earlier in the book: $$d(f,g) = \sum_{k=1}^{\infty} 2^{-k}\frac{d_k(f,g)}{1+d_k(f,g)}$$ where $$d_k(f,g) = \max_{\lvert x\rvert \leq k}\lvert f(x) - g(x)\rvert$$
The proof would then go as follows:
Let $f_n$ be a Cauchy sequence in $C(\mathbb{R})$.
Now, for any $\alpha \in \mathbb{N}$, suppose $f_n$ was not Cauchy on $[-\alpha, \alpha]$ under the supremum norm. Then there exists an $\epsilon > 0$ such that for all $N \in \mathbb{N}$ there exists an $n, m > N$ such that $\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f_m(x)\rvert\geq \epsilon$. Then we would have that with this $\epsilon$, for all $N \in \mathbb{N}$ there exists an $n, m > N$ such that \begin{align*}d(f_n,f_m) = \sum_{k=1}^{\infty} 2^{-k}\frac{d_k(f_n,f_m)}{1+d_k(f_n,f_m)} \geq 2^{-\alpha}\frac{d_{\alpha}(f_n,f_m)}{1+d_{\alpha}(f_n,f_m)} &= 2^{-\alpha}\frac{\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f_m(x)\rvert}{1+\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f_m(x)\rvert}\\ &\geq 2^{-\alpha}\frac{\epsilon}{1+\epsilon} > 0\end{align*}
The last inequality is due to $\frac{x}{1+x}$ being an increasing function.
Thus $f_n$ would not be Cauchy in $C(\mathbb{R})$. So we must have that $f_n$ is Cauchy on $[-\alpha, \alpha]$ under the supremum norm for any $\alpha \in \mathbb{N}$.
It can be shown that $C[-\alpha, \alpha]$ is complete. Since you are already working in Carothers, if you need help proving this the proof is almost identical to Lemma 10.8. Since $C[-\alpha, \alpha]$ is complete and $f_n$ is uniformly Cauchy, $f_n$ converges uniformly to some function $f_{\alpha}$ in $C[-\alpha,\alpha]$. Since $f_{\alpha}$ agrees with $f_{\beta}$ on $[-\beta,\beta]$ for every $\beta < \alpha$, we have that $f = \lim_{\alpha \rightarrow \infty} f_{\alpha}$ is a well defined continuous function on $\mathbb{R}$.
So, let $\epsilon > 0$ and choose an $\alpha$ so large that $2^{-\alpha} < \frac{\epsilon}{2}$ and a $n$ so large that $\max_{\lvert x\rvert < \alpha} \lvert f_n(x) - f(x)\rvert < \frac{\epsilon}{2}$ as $f_n$ converges uniformly to $f$ on $[-\alpha, \alpha]$.
Then we have that:
\begin{align*}d(f_n, f) = \sum_{k=1}^{\infty} 2^{-k}\frac{d_k(f_n,f)}{1+d_k(f_n,f)} &= \sum_{k=1}^{\alpha} 2^{-k}\frac{d_k(f_n,f)}{1+d_k(f_n,f)} + \sum_{k=\alpha+1}^{\infty} 2^{-k}\frac{d_k(f_n,f)}{1+d_k(f_n,f)}\\ &\leq \sum_{k=1}^{\alpha}\frac{\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f(x)\rvert}{1+\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f(x)\rvert} + \sum_{k=\alpha+1}^{\infty}2^{-k}\\ &\leq \sum_{k=1}^{\alpha}2^{-k}\max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f(x)\rvert + 2^{-\alpha}\\ &\leq \max_{\lvert x\rvert \leq \alpha}\lvert f_n(x) - f(x)\rvert + 2^{-\alpha} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\end{align*}
