Two sequences are subsequences of one another, one converges. Are they the same? Let $a_n$ converge to a. Let $b_n$ be a subsequence of $a_n$ and $a_n$ be a subsequence of $b_n$. Are they the same?
I've tried showing they are by contradiction, the Bolzano-Weierstrass theorem seems relevant but I can't see where. I can't see any counter example but there seems to be some pattern to why examples fail but I can't see it.
Removing the restriction of $a_n$ converging seems to allow for counter examples which all seem to alternate, if I could prove that they all alternate then it would be done as alternating series don't converge.
Hints would be most welcome!
 A: Suppose for a contradiction that $a_1 \ne b_1$. Since $a_n$ is a subsequence of $b_n$, there exists a $c_1>1$ s.t. $a_1=b_{c_1}$. For each $i=\{1,...,c_1\}$ there exists $n_i>i$ s.t. $a_{n_i}=b_i$, because $b_n$ is a subsequence of $a_n$. Let $c_2=n_{c_1}$. Now, for each $i=\{1,...,c_2\}$ there exists $n_i>i$ s.t. $b_{n_i}=a_i$. Let $c_3=n_{c_2}$. And so on.
Note that $1<c_1<c_2<c_3<...$ and also that $a_1=b_{c_1}=b_{c_3}=...$.
So $b_n$ has a convergent subsequence with limit $a_1$.
Similarly (starting at $b_1$) we get that $b_n$ has a convergent subsequence with limit $b_1$. That contradicts $a_1 \ne b_1$, since $b_n$ is convergent as well. So we have $a_1=b_1$.
Now considering the sequences starting at $a_2$ and $b_2$ and so on, we get that $a_n=b_n$ for all $n$.
A: Definition A sequence $b_n$ is said to be a subsequence of a sequence $a_n$ iff there exists a strictly increasing function $\phi\ :\ \mathbb{N}\rightarrow\mathbb{N}$ such that $b_n=a_{\phi\left(n\right)}$ for each $n\in\mathbb{N}$.


Proof Let $a_n,b_n$ be a subsequence of each other. Then, there exist strictly increasing functions $\phi,\varphi\ :\ \mathbb{N}\rightarrow\mathbb{N}$ such that $a_n=b_{\phi(n)}$ and $b_n=a_{\varphi(n)}$ for every $n\in\mathbb{N}$. For every $n\in\mathbb{N}$,
$$a_n=b_{\phi(n)}=a_{\varphi(\phi(n))} \ \mathrm{and}\  b_n=a_{\varphi(n)}=b_{\phi(\varphi(n))}.$$
From this, we get $\varphi\circ\phi=I=\phi\circ\varphi$; therefore we write $\varphi=\phi^{-1}$. What this says to us is that the functions $\varphi$ and $\phi$ are bijective.


Assume that $\varphi$ is not the identity function on $\mathbb{N}$, that is, there exists a natural number, say $n_0$, such that $\varphi(n_0)\neq n_0$. Set $\mathbb{N}_x=\{n\in\mathbb{N}\ |\ n<x\}$. It follows from $\varphi(n_0)\neq n_0$ that the sets $\mathbb{N}_{n_0}$ and $\mathbb{N}_{\varphi(n_0)}$ have finite but different cardinalities. It is well-known that the function $\varphi^*\ :\ \mathbb{N}_{n_0}\rightarrow\varphi(\mathbb{N}_{n_0})$ defined by
$$\varphi^*(n)=\varphi(n)\mathrm{\ for\ each\ }n\in\mathbb{N_{n_0}}$$ is bijective. Since the function $\varphi$ is a strictly increasing,  we have \begin{array}{rcl}
\varphi(\mathbb{N}_{n_0})&=&\{\varphi(n)\in\mathbb{N}\ |\ n<n_0\}\\
&=&\{\varphi(n)\in\mathbb{N}\ |\ \varphi(n)<\varphi(n_0)\}\\
&=&\mathbb{N}_{\varphi(n_0)}
\end{array} which contradicts the sets $\mathbb{N}_{n_0}$ and $\mathbb{N}_{\varphi(n_0)}$ having finite but different cardinalities. Thus, $\varphi=I$ and so $(a_n)=(b_n)$.
