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If $(a_n)$ is subsequence of $(b_n)$ and $(b_n)$ is a subsequence of $(a_n)$, can we conclude that $(a_n) = (b_n)$?

Is their a way I can disprove this? I don't know how I would go about creating two subsequence that are subsequences of one another, and show that they aren't equal.

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    Try any two sequences of $0$s and $1$s, with infinitely many $0$s and infinitely many $1$s. – 2017-02-20
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    if I let an = {0,1,0,1,0...} and bn = { 1, 0, 0, 1, 0, 0, ...} would that work? would those two be equal? – 2017-02-20
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    What does it mean for two sequences to be equal? Is each of those a subsequence of the other? – 2017-02-20
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    thats what Im confused on I don't know how to tell.. – 2017-02-20
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    If $f:X\to Y$ and $g:X\to Y$ are functions, then $f=g$ if (by definition) for all $x\in X$, $f(x)=g(x)$. Sequences are functions whose domain is $\mathbb N$. For sequences $(a_n)$ and $(b_n)$ to be equal means for all $n$, $a_n=b_n$. For the examples you gave, check if they are equal. Also, try to see how $(b_n)$ can be expressed as a subsequence of $(a_n)$ and vice versa. – 2017-02-20

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For a simple example of such a pair of sequences that are not equal:

$$\begin{align} \{a_n\} = \{2^n \bmod 11\} &= \{1,2,4,8,5,10,9,7,3,6,1,2,4,\ldots\}\\ \{b_n\} = \{7\cdot2^n \bmod 11\} &= \{7,3,6,1,2,4,8,5,10,9,7,3,6,\ldots\}\\ \end{align}$$

Any such pair of sequences that are non-equal mutual subsequences must have some kind of repetitive nature, if not necessarily cycling of this simple form, since we also have that $\{a_i\}$ is a proper subsequence of itself (and likewise $\{b_i\}$).

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    @JonasMeyer good point. updated to soften that conclusion. – 2017-03-06