I wanted to prove that the space of continuous functions and also the space of differentiable functions are not complete but I don't know how to construct a Cauchy sequence that doesn't converge to these spaces?
The metric of the continuous function is the sup metric and the metric of the metric of the differentiable functions is:
$\sum_{|\alpha|\leq k}$max|D$^\alpha$f(x) - D$^\alpha$g(x)|
where D$^\alpha$ is the multi-index notation. The functions of both spaces are defined on a bounded domain of R$^n$.
Another question is that "Where do exactly these sequences come from?", I mean when I faced a problem concerning the completeness of a space, "How should I construct such a sequence?".
Is there a special "tip" for that?
Thank you so much for sharing your experience.
If the set is complete, you need to show that the limit of any Cauchy sequence must be a member of the set. If the set is not complete, you need to find a Cauchy sequences that converges to a limit outside of the set.
For the continuous case, the first approach will work. It looks like you are looking at the set of continuous functions from some bounded subset $S\in\mathbb R^n$ to $\mathbb R$, using the sup norm:
$$||f-g||=sup\{|f(\mathbf x)-g(\mathbf x)|,\ \mathbf x\in S\}$$
We then let $f_n:S\to\mathbb R$ be s Cauchy sequence of continuous functions.
$$\forall \epsilon>0,\ \exists N,\ s.t.\ n,m>N\implies ||f_n-f_m||<\epsilon$$
choosing a particular $\mathbf x$:
$$\forall \epsilon>0,\ \exists N,\ s.t.\ n,m>N\implies |f_n(\mathbf x)-f_m(\mathbf x)|<\epsilon$$
This means that $f_n$ coverges pointwise. Since $\mathbb R$ is complete, we know there is a function $f:S\to\mathbb R$ such that $f(\mathbf x)=\lim_{n\to\infty}f_n(\mathbf x)$ and that $f_n$ converges uniformly to $f$:
$$\forall \epsilon>0,\ \exists N,\ s.t.\ n>N\implies ||f_n-f||<\epsilon$$
We now need to show $f$ is continuous, starting from the statement that every $f_n$ is continuous:
$$\forall \mathbf x\in S,\ \forall \epsilon>0,\ \exists \delta\ s.t.\ ||\mathbf y-\mathbf x||<\delta\implies |f_n(\mathbf y)-f_n(\mathbf x)|<\epsilon$$
Here $||\mathbf y-\mathbf x||$ indicates a euclidean distance. We can now construct the contiuity statement for $f$. Fix $\epsilon>0$ and $\mathbf x\in S$.
$$\exists n\ s.t.\ ||f_n-f||<\frac\epsilon 3\implies\ sup\{|f_n(\mathbf x)-f(\mathbf x)|,\ x\in S\}<\frac\epsilon 3$$
$$\exists \delta\ s.t.\ ||\mathbf y-\mathbf x||<\delta\implies|f_n(\mathbf y)-f_n(\mathbf x)|<\frac\epsilon 3$$
By triangle inequality:
$$|f(\mathbf y)-f(\mathbf x)|\le|f(\mathbf y)-f_n(\mathbf y)|+|f_n(\mathbf y)-f_n(\mathbf x)|+|f_n(\mathbf x)-f(\mathbf x)|$$
The inequalities above force each of these terms to be strictly less than $\frac\epsilon 3$. Therefore the $\delta$ chosen above is sufficient to say:
$$||\mathbf y-\mathbf x||<\delta\implies|f(\mathbf y)-f(\mathbf x)|<\epsilon$$
Which by definition means the limit function $f$ is continuous. Since this can be applied to any Cauchy sequence of continuous functions, the set is complete.
I'm not familiar with the notation for the metric for differentiable functions, so I can't say anything about completeness. If it is complete, a similar proof will probably be needed. If it is not complete, you can choose a familiar function that is not everywhere differentiable like $f(\mathbf x)=||\mathbf x||$ and find a way to approach it with differentialbe functions so that the norm goes to $0$.