I've seen and read the various asked questions about fixed points posted here previously. In the course I'm doing I haven't defined contraction mappings or even a fixed point directly. All I have at my disposal are the ideas of continuity and convergent sequences. As such here is how the question was framed:
Suppose $f: \mathbb{R}^{n} \to \mathbb{R}^{n}$ is continuous and $\mathbf{x_{0}}$ is arbitrary. Define a sequence $\mathbf{x_{k}} = f(\mathbf{x_{k-1}})$. Prove that if $\mathbf{x_{k}} \to \mathbf{a}$ then $f(\mathbf{a}) = \mathbf{a}$. We say $\mathbf{a}$ is a fixed point of $f$.
My attempt:
Given that $f$ is continuous means that for all $\epsilon > 0$ there is a $\delta >0$ such that if $\|\mathbf{x - a}\| < \delta$, then $\|f(\mathbf{x_{k}}) - f(\mathbf{a})\| < \frac{\epsilon}{2}$. As well given that $\mathbf{x_{k}} \to \mathbf{a}$, then for all $\epsilon > 0$, there is a $K>0$ such that if $k > K$, then $\|\mathbf{x_{k} - a}\| < \frac{\epsilon}{2}$.
Choose $\delta > 0$ and $K$ that satisfy these two conditions.
Observe:
$$\|\mathbf{f(a) - a}\| \leq \|\mathbf{f(a) - f(x_{k})}\| + \|\mathbf{f(x_{k}) - a}\| = \|\mathbf{f(a) - f(x_{k})}\| + \|\mathbf{x_{k + 1} - a}\| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$
I substituted $\mathbf{x_{k+1}}$ in fro $\mathbf{f(x_{k})}$ based on the definition of the function.
Therefore $\mathbf{f(a) = a}$.
The one thing I'm concerned about is having a bit of an idea of what a fixed point is supposed to mean.....so in the form that I proved this are we saying that it is only after a certain point in the sequence that the point will become a fixed point of the function? In terms of the ideas of sequences I would gather this is the case as once a sequence converges to a point every itteration after that is the point it converged to. Is that the same idea going on here?
You only need the sequential characterisation of continuity:
$$f(a) = f(\lim_{k \to \infty} x_k)= \lim_{k\to \infty} f(x_k) = \lim_{k\to \infty} x_{k+1} = a$$