Sequence of points in metric space

Question

Let $(x_n)_n$ be a sequence such that for every $n$ $$d(x_{n+1},x_n) \le \frac 1 3 d(x_n,x_{n-1})$$ I am trying to show that this implies that for every $n$ and $m$ $$d(x_{m+1},x_{n+1}) \le \frac 1 2 d(x_m,x_n)$$ I am struggling to do this though. I am guessing the $1/2$ comes from the fact that $1/3 + 1/3^2 + 1/3^3 + \cdots = 1/2$. I have tried using the triangle inequality to expand out the LHS so that I can apply the first inequality but I am not sure how to get it in terms of $d(x_m,x_n)$. Hints would be most appreciated.

Answer 1

This seems false. Let $X$ be the metric space $\mathbb{R}$ with the Euclidean metric. Take $x_0 = 1$, $x_1 = 0$, $x_2 = \frac{1}{3}$ and $x_3 = \frac{4}{9}$. Then we have $$ d(x_{n + 1}, x_n) \leq \frac{1}{3} d(x_n, x_{n - 1}) $$ for $n = 1, 2, 3$. However, we have $$ d(x_3, x_1) = \frac{4}{9} \nleq \frac{1}{2} d(x_2, x_0) = \frac{1}{3}. $$