Label $X$ as the real interval $[0, a]$ where $a \in \mathbb{R}^+$, so that $\text{int}(X) = (0, a)$ labels the interior of $X$ and $\partial X$ labels the boundary of $X$. I have a function $f:\text{int}(X) \rightarrow \text{int}(X)$ that is well defined on the interior of $X$, but ambiguous on the boundary of $X$; that is $f(\partial X)$ is not a defined mapping action.
Further, the system has the following properties: $f$ is a contraction mapping over $\text{int}(X)$, and if $a$ is not a positive integer, then sequential application of $f$ over any point $x \in \text{int}(X)$ will converge to a fixed point in $\text{int}(X)$. However if $a \in \mathbb{N}^+$, then application of $f$ over any point $x \in \text{int}(X)$ will converge to a boundary value in $\partial X$.
As such, I have a fixed map $f$ defined over a fixed interval $(0, a)$ where the limiting behavior of $f$ over initial point $x \in (0, a)$ is determined by parameter $a$. If $a$ is a positive integer, iteration of $f$ will converge to either 0 or $a$, where $f$ is not defined.
I want to prove that sequential application of $f$ over $x \in \text{int}(X)$ will converge to a fixed point in $\text{int}(X)$ if $a \in \mathbb{R}^+ \setminus \mathbb{N}^+$. I am wondering if I can use the contraction mapping theorem to achieve this aim. After all, the contraction mapping theorem requires that the space is complete, however if action of $f$ is only defined over $(0, a)$ rather than $[0, a]$ then the space is not complete and the theorem cannot be invoked.
Does anyone know how I can then adequately set up this problem to eventually prove that iteration of $f$ over $\text{int}(X)$ will converge to a fixed point in $\text{int}(X)$ if $a \in \mathbb{R}^+ \setminus \mathbb{N}^+$ using the contraction mapping theorem or something else?
Edit: Consider the following Theorem (Topology, Gamelin, Green) which will allow me to reword the above in a different way.
Theorem: Let $X$ be a complete metric space with metric $d$, and let $S$ be a metric space. Let $c$ be a fixed constant with $0 < c < 1$. Suppose that $(s, x) \rightarrow \Phi_s(x)$ is a continuous function from $S\times X$ to $X$ such that $$ d(\Phi_s(x), \Phi_s(y)) \leq cd(x, y) $$ for $x, y \in X, s \in S$. Then for each $s \in S$, there is a unique point $x_s^* \in X$ such that $\Phi_s(x_s^*) = x_s^*$. Furthermore, $x_s^*$ depends continuously on $S$.
Now, given the above theorem, we can let $a$ be the continuous parameter $s$ as stated in the above theorem, and I am now studying a function $\Phi_a: (0, a) \rightarrow (0, a)$ where the fixed point $x_a^*$ does not depend continuously on $a$, but rather depends piecewise discontinuously on $a$, and clearly $(0, a)$ is not a complete metric space as required.
As such, If I was to first establish that iteration of $\Phi_a$ is undefined when $a \in \mathbb{N}^+$, could I then apply the above theorem for all $a \in \mathbb{R}^+ \setminus \mathbb{N}^+$ and put $\Phi_a: [0, a] \rightarrow [0, a]$ so that we have a complete metric space and the behavior of the map is well defined for all such values of $a$?
