Let $f:\mathbb{R}^{n} \mapsto \mathbb{R}^{n}$, given any $x_{0} \in \mathbb{R}^n$, we can generate a sequence such that $x_{k} = f(x_{k-1})$ for $k \geq 1$. If for some $a$ , we have $\lim\limits_{ x \to a }f(x) = a$, does there exist some open neighborhood $N$ around $a$ and $x_{0} \in N$, such that $\lim\limits_{ k \to \infty }x_{k} = a$?
I am curious about this because I kind of want to take the limit on both side of $x_{k} = f(x_{k-1})$, and solve for the limits. However, this only works if we know $x_{k}$ is convergent in the first place, so I'm wondering about this case.
This statement is automatically true if $f$ is continuous at $a$, as we can set $x_{0} = a$. Also if we instead say $x_{0} \in N \setminus \{ x_{0} \}$, then taking f as the identity would make the statement false. However, I wonder in the stated way, is this true in general?
This is probably related to some dynamical system concepts, but unfortunately I don't have much experience in this area besides a basic course on ODE. I have looked at examples where a point is attractive and not stable, but that also did not lead me anywhere.
Here is a counter-example: $$f(x)=\left\{\begin{matrix} 2\cdot x & \text{when } x \ne 0 \\ 1 & \text{when } x=0 \end{matrix}\right.$$ for $a=0$. For any vicinity around $0$ and any initial $x_0$ in the vicinity, the sequence generated by $x_{k}=f(x_{k-1})$ will, eventually, "escape" the vicinity.