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How to find the cardinality of the set of divergent sequences? Let's name this set $A$.

I know that cardinality of the set of sequences equals $2^{\lvert \mathbb{N}\rvert}$, so $\lvert A\rvert\le 2^{\lvert \mathbb{N}\rvert}$. How to strictly prove to the other side?

Thank you for your time.

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up vote 3 down vote accepted

Pick one divergent sequence, $a_n$, now for every $r\in\Bbb R$ let $r_n=\begin{cases} r & n=1\\ a_n & n>1\end{cases}$.

Show that this is an injection from $\Bbb R$ into $A$.

Alternatively, pick one sequence whose elements are pairwise different (e.g. $a_n=n$), and for every $K\subseteq\Bbb N$ define the sequence: $k_n=\begin{cases} a_n & n\in K\\ -a_n & n\notin K\end{cases}$

Show that this is an injection from $\mathcal P(\Bbb N)$ into $A$.

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+1 that is a very neat solution and seems so easy. – Dominic Michaelis May 26 '13 at 8:47
Dominic, thank you. It was hard to me to just leave a hint, with such a short solution. – Asaf Karagila May 26 '13 at 8:51
but i can not understand this solution, can you show more details? – kushtibargo May 26 '13 at 8:53
I think every Hint that leads to this solution will be longer than the solution itself. – Dominic Michaelis May 26 '13 at 8:54
@kushtibargo Well do you unterstand that every constructed sequence is divergent? And how many are there? As many as you have real numbers. That makes $\mathfrak{c}$ many. And as $|2^\mathbb{N}|=\mathfrak{c}$ we are done here. – Dominic Michaelis May 26 '13 at 8:55

Hint: Show that the set of unbounded increasing sequences has cardinality $\mathfrak{c}$.

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For any non-rational $\,r\in\Bbb R\,$ , the sequence $\;\{a_{r,n}\}_{n\in\Bbb N}:= \{rn\}_{n\in\Bbb N}\;$ is divergent

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doesn't that work for every $r\neq 0$ ? – Dominic Michaelis May 26 '13 at 8:46
Yes @DominicMichaelis, yet on purpose (as part of the hint) I wanted the OP to focus on irrational ones. – DonAntonio May 26 '13 at 9:26

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