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If $(x_n)$ is a sequence in $\mathbb R$ such that for each $m \in\mathbb N$ with $m > 1$,  the subsequence $(x_{mn})$ of $(x_n)$ is convergent, then can we conclude that $(x_n)$ is also  convergent?
in Real Analysis by Professor | 328 views

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No. We can't conclude anything. Note that for each $m,n>1$, $mn$ is always a composite number. So we can define a sequence which takes a constant value $c_1$  on each composite number and another constant value $c_2$ on prime numbers. Then we can see that each of its subsequence of the form $(x_{mn})$ will converge to $c_1$ whereas the subsequence $(x_p)$ where $p$ runs over the prime number sets then $(x_p)$ will converge to $c_2$. This conclude that our sequence is not convergent.
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