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I'm looking at this exercise solution and there is a last step which I do not really understand. Consider the set of continuous functions on the interval $X$, that is

$C(X):=􏰁\{f:X→R \mid f \ \text{is continuous on}\ X􏰂\}$

with the supremum norm $\mid\mid\cdot\mid\mid_\infty$. We want to prove the completeness of the space. So we take an arbitrary Cauchy-sequence which is also Cauchy if evaluated at $x\in X$ and also convergent by the completeness of $\mathbb{R}$. So at this point we can define the limit of the evaluated sequence of function and check that the limit is a function lying in our space $C(X)$ and that indeed our sequence of functions converges to this function. To prove that this function lies indeed in $C(X)$ the argument given is the following one:

$\lim_{n\rightarrow \infty} f_n(x) = f(x)$, for each $x \in X$.

$d(f_n,f) = \sup_{x\in X} \mid f_n(x)−f(x)\mid=\sup_{x\in X} \lim_{m\rightarrow \infty}\mid f_n(x)−f_m(x)\mid$

$\leq\liminf_{m\rightarrow \infty}\sup_{x\in X} \mid f_n(x)−f_m(x)\mid\leq\epsilon$

The exercise was made considering $X=[0,1]$ but I would like to generalize. My problem is essential the last step passing to liminf. I really don't see why we are allowed to do this.

Thank you.

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You probably want to assume that $X$ is compact so the supremum over $X$ will exist for all functions. Here's an argument that avoids $\liminf$:

Take $\{f_n\}_{n \geq 0}$ a $\|\cdot\|_\infty$-Cauchy sequence in $C(X)$. Let $\epsilon > 0$. There exists $n_0 > 0$ such that: $$\begin{align} \|f_n-f_m\|_\infty &< \epsilon,\quad \forall\,m,n>n_0 \\ \sup_{x \in X}|f_n(x)-f_m(x)| &< \epsilon,\quad\forall\,m,n>n_0 \\ |f_n(x)-f_m(x)|&<\epsilon,\quad \forall\,m,n>n_0,\quad\forall\,x\in X \end{align}$$So fixed $x$, $\{f_n(x)\}_{n \geq 0}$ is a Cauchy sequence in $\Bbb R$ (or $\Bbb C$, works too), and hence it converges. Denote $f(x) = \lim_n f_n(x)$. Now we prove that $f_n \stackrel{\|\cdot\|_\infty}{\longrightarrow}f$. Let $\epsilon > 0$. There exists $n_0 > 0$ such that: $$\begin{align} \|f_n-f_m\|_\infty &< \epsilon,\quad \forall\,m,n>n_0 \\ \sup_{x \in X}|f_n(x)-f_m(x)| &< \epsilon,\quad\forall\,m,n>n_0 \\ |f_n(x)-f_m(x)|&<\epsilon,\quad \forall\,m,n>n_0,\quad\forall\,x \in X \\ \lim_{m \to +\infty}|f_n(x)-f_m(x)| &\leq \epsilon,\quad\forall\,n>n_0,\quad \forall\,x \in X \\ |f_n(x)-f(x)|&\leq \epsilon,\quad\forall\,n>n_0,\quad \forall\,x \in X\\ \sup_{x \in X}|f_n(x)-f(x)|&\leq \epsilon,\quad\forall\,n>n_0 \\ \|f_n-f\|_\infty &\leq \epsilon,\quad\forall\,n>n_0\end{align}$$

Now, we prove that $f$ is continuous at an arbitrary $x_0 \in X$. Let $\epsilon > 0$. There exists $n_0 > 0$ such that $\|f_{n_0}-f\|_\infty < \epsilon/3$, by the above work. This particular $f_{n_0}$ is continuous, so there is a neighborhood $V$ of $x_0$ such that $x \in V \implies |f_{n_0}(x)-f_{n_0}(x_0)|<\epsilon/3$. So $x \in V$ gives: $$|f(x)-f(x_0)| \leq |f(x)-f_{n_0}(x)|+|f_{n_0}(x)-f_{n_0}(x_0)|+|f_{n_0}(x_0)-f(x_0)| < \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} = \epsilon.$$

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  • $\begingroup$ That's a nice way! Thanks. $\endgroup$ – sky90 Jun 7 '15 at 6:12

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