recursive series in a metric-space Let's say that you have a series in a metrix space, defined recursively with $x_{n+1} = f(x_n)$.
Let's also say that you are given the knowledge that this series converges.
Is it then possible that f is not continous?
(The reason I am asking is that I read two different proofs of Banach's fixed point theorem. In one proof he used that since the contraction f is continious and he knows that the series converges, he gets
$a=lim_{n\to \infty} x_{n+1}= lim_{n\to \infty}f(x_n) \text{---continuity-->} = f( lim_{n\to\infty}x_n)  = f(a)$.
But in the other proof, he only says that since he knows that the series converges, he onlys says that we must have $a=f(a)$.)
So this makes me wonder, if we are given info that $x(n+1)= f(x_n)$, converges, without nececarrily knowing that f is continous, are there then examples of functions where f is not continious and this work, or can we deduce from this that $f$ must be continous? And you still have that $f(a)=a$?
 A: Let $X = [0,1]$ and
$$f(x) = \begin{cases} 1 &, x = 0\\ \frac{x}{2} &, x \neq 0. \end{cases}$$
For every $x_0 \in X$, the sequence defined by the recursion converges (to $0$), yet $f$ is not continuous and has no fixed point.
The variant
$$g(x) = \begin{cases} 1 &, x \in X\setminus \mathbb{Q}\\ \frac{x}{2} & x \in X\cap\mathbb{Q} \end{cases}$$
gives an example of a function that has a unique fixed point, such that the sequence obtained by iteratively applying $g$ converges to the fixed point for all starting points, but the function is not continuous in the fixed point (nor in any other point, here). The continuity of the function cannot be deduced from the result that every iterated sequence converges to the unique fixed point, it must be - in one form or other - part of the premises.
In the Banach fixed point theorem, one of the premises is the existence of a $q < 1$ such that for all $x,y$ we have $d(f(x),f(y)) \leqslant q\cdot d(x,y)$. That premise implies the continuity of $f$, and was apparently implicitly used in the second proof.
