I hope this question is appropriate for this site, I was unable to get an answer on Math StackExchange.  

Label $X$ as the real interval $[0, a]$ where $a \in \mathbb{R}^+$, so that $\text{int}(X)$ 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 defined.  

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 $\partial X$.

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 $a \in \mathbb{R}^+ \setminus \mathbb{N}^+$, however it is unclear to me 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 $\text{int}(X)$ then $\text{int}(X)$ is decidedly not a complete space.  

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 $a \in \mathbb{R}^+ \setminus \mathbb{N}^+$, and that the limiting behavior of $f$ is undefined otherwise?