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Let $(x_n)$ be a sequence of real numbers such that $\lim_{k \rightarrow \infty} x_{2k}=a$ and $\lim_{k \rightarrow \infty}x_{2k+1}=b$. Prove that the set of subsequential limit of $x_n$ is the set $\{a,b\}$.

I know that if $x_{n_k}$ is a convergent subsequence, then infinitely many of them are either odd or even, but i dont know how to show that it converges to either a or b. thank in advance.

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Hint: For any $c$ with $c \ne a$ and $c \ne b$ put $\varepsilon := \dfrac{\min\{|c-a|, |c-b|\}}{2}$ and consider the $\varepsilon$-neighborhoods of the points $a$, $b$, and $c$.

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  • $\begingroup$ sorry, can you say more... i couldnt get it... $\endgroup$ – gerry lang Oct 24 '13 at 16:12
  • $\begingroup$ There is $N_\varepsilon$ such that for all $n > N_\varepsilon$ we have $x_{n} \in (a-\varepsilon, a+\varepsilon) \cup (b-\varepsilon,b+\varepsilon)$. Hence there may only be a finite number of elements in the $\varepsilon$-neighborhood of $c$. $\endgroup$ – njguliyev Oct 24 '13 at 16:16

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