Let $\{x_n\}$ be a sequence of real numbers in the interval $[0, 3/2]$. Suppose that $\lim_{n \rightarrow \infty} |x_n-x_{n+1}|=1$. Prove that no subsequence of $\{x_n\}$ converges to $3/4$.

My inclination is to prove this by contradiction. Suppose, to the contrary, that there exists a subsequence $\{x_{n_k}\}$ such that $x_{n_k}\rightarrow 3/4$. Then for every $\epsilon >0$ there exists an $N$ such that $n_k\geq N$ implies $|x_{n_k}-3/4|<\epsilon$. Then for all $n_k \geq N$, $$-\epsilon +3/4 \leq x_{n_k} \leq 3/4 + \epsilon.$$

Not sure what to do from here. I also tried using that convergent implies Cauchy, but I was also not sure what to do. Any help is appreciated, thank you.


If there is such a subsequence, then there are terms very close to 3/4. But since $\lim_{n \to \infty}|x_n - x_{n+1}| = 1$, there must be terms very close to $3/4 \pm 1$. This would be a contradiction because $x_n \in [0,3/2]$

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    $\begingroup$ Nicely said, but OP should be aware that this is (at least in part) an informal argument and there are details to fill in. $\endgroup$ – Wojowu Oct 31 '16 at 15:04

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