Let $(x_n)$ and $(y_n)$ be sequences such that $(x_n)$ is a subsequence of $(y_n)$ and
$(y_n)$ is a subsequence of $(x_n)$. Given that $x_n$ converges, does it follow that $x_n=y_n$ for all $n$?
I strongly suspect this is true. I have an idea but I'm not convinced its super tight.
By our subsequence fact, let $x_{a_n}=y_n$ and $y_{b_n}=x_n$. Since we have to fit these terms in the sequence and they are all in order, we must have $a_n\ge n$ and $b_n \ge n$ with equality if and only if the first $n-1$ terms are identical.
Assume BWOC $\exists k: x_k\neq y_k$ with $k$ minimal. Well $x_{a_k}=y_k$ with $a_k\gt k$, but $y_{b_{a_k}}=x_{a_k}, b_{a_k}\gt a_k \ (>k)$. In other words, $x_k \neq y_k \implies \text{$y_k$ appears later in the sequence ($x_n$)}\implies \text{another $y_k$ appears later in the sequence $(y_n)$}$ and this keeps going on back and forth. Since $(x_n)$ converges and $y_k$ appears infinitely many times in $(x_n)$, it must converge to $y_k$. But, by the same argument, we can show it converges to $x_k$. Since $x_k\neq y_k$, this is a contradiction.
After writing this out, I feel like my idea is right but I just have this doubt in my execution that I can't shake. Is my argument correct or is the proposition false after all?
Best Answer
Your statement and the idea for the proof is correct. I would write is as follows, using the notation $a(n)$, $b(n)$, ... instead of $a_n$, $b_n$, ... to avoid deeply nested indices:
$(x_n)$ is a subsequence of $(y_n)$, so there is a strictly increasing function $a: \Bbb N \to \Bbb N$ with $x_{a(n)} = y_n$ for all $n$.
Similarly, $(y_n)$ is a subsequence of $(x_n)$, so there is a strictly increasing function $b: \Bbb N \to \Bbb N$ with $y_{b(n)} = x_n$ for all $n$. In particular, $x_{a(b(n))} = x_n$ for all $n$.
Assume that the sequences are not identical, and define $k = \min \{ n \mid x_n \ne y_n \}$. Then $a(k) > k$ and $b(k) > k$, which implies $a(n) > n$ and $b(n) > n$ for all $n \ge k$.
Now define two functions $c, d: \Bbb N \to \Bbb N$ recursively as $$ c(0) = k \, , \, c(j+1) = a(b(c(j))) \, ,\\ d(0) = a(k) \, , \, d(j+1) = a(b(d(j))) \, . $$ Both functions are strictly increasing. Then $$ x_k = x_{c(0)} = x_{c(1)} = x_{c(2)} = \cdots $$ and $$ y_k = x_{d(0)} = x_{d(1)} = x_{d(2)} = \cdots $$ Since $(x_n)$ is convergent, all its subsequences have the same limit, and it follows that $$ x_k = \lim_{n \to \infty} x_n = y_k \, . $$ That is a contradiction to the construction of $k$.