Consider the metric space C[a,b] equipped with the metric $d(f,g)=max|f(x)-g(x)|$. I would like to show that the sequence $((x/c)^n)$ has no convergent subsequence, where c is a constant such that $x/c \in [0,1] $. My attempt:
Assume it does, then there is a continuous function $f$ such that a subsequence $((x/c)^{n_k})$ converges to it. Then $((x/c)^{n_k})$ cauchy. Thus for all $\epsilon \gt 0$ there exists K such that for any $k_n, k_m \gt K$, $d(f_{k_n},f_{k_m}) \lt \epsilon$. Then $max_{x \in[a,b]}|x^{k_n}/c^{k_n}-x^{k_m}/c^{k_m}|\lt \epsilon$. I would like to derive a contradiction from this, perhaps using the fact that $(x/c)^n$ is not uniformly continuous in the reals. However, I am uncertain of how to proceed.
Any help is appreciated
This is false. Take $a=0,b=1$. If $c$ is large enough then $\sup_x |\frac x c| <\frac 1 2$ so the entire sequence $(\frac x c)^{n}$ converges uniformly to $0$.
If your question is to find a $c$ such that no subsequence of $(\frac x c)^{n}$ converges uniformly you can do the following: suppose $0<a<b$. take $c=b$. Then $(\frac x c)^{n}$ converges to $0$ at each point $x \in [a,b)$ and to $1$ at $x=b$. Since the limitimg function is not continuous it follows that no subsequence of $(\frac x c)^{n}$ converges uniformly. I will let you handle the other cases.