Let $(X,d)$ be a metric space. Suppose there is a sequence $(x_n)$ in $X$ such that $(x_{2n})$ is Cauchy and $d(x_n,x_{n+1})\to 0$. The question is whether $(x_n)$ is Cauchy?
It seems intuitively that this is false, for there might be a sequence whose even terms come arbitrarily close to each other and the subsequent terms come close together as well but any two far off odd terms will still be separated for the consecutive small distances might add up to a large value.
Can someone through any light?
Since $(x_{2n})_{n\geq 1}$ by definition of Cauchy sequence we have that for a give $\epsilon>0$ there is a $n_0$ such that for $n,m>n_0$ we have $d(x_{2n},x_{2m})\leq \frac{\epsilon}{3}$. An by definition of convergence to zero, we have that given $\epsilon$ there is $n_1$ such that $n>n_1$ $d(x_n,x_{n+1})<\frac{\epsilon}{3}$ . Now, supose $m,n >m_0= \max \{n_0,N_1\}$, then we have three possiblities: m,n are evem numbers $m=2k, n=2s$, then $$d(x_{m},x_{m})=d(x_{2k},x_{2s})\leq \frac{\epsilon}{3} < \epsilon$$ $m=2k$ is even and $n =2s+1$is odd, so by triangular inequality $$d(x_m,x_n)\leq d(x_m,x_{n+1}))+d(x_{n+1},x_n)= d(x_{2k},x_{2t+2}))+d(x_{n+1},x_n)\leq \frac{2\epsilon}{3}<\epsilon $$ or $m=2k+1$, $n=2t+1$, and using the previous estimate $$d(x_m,x_n)=d(x_{2k+1},x_{2t+1})\leq d(x_{2k+1},x_{2t+2})+d(x_{2t+2},x_{2t+1})\leq \frac{2\epsilon}{3}+\frac{\epsilon}{3}=\epsilon $$ Therefore, it is a Cauchy sequence.