Rearrangement of Sequence $\left\{x_n\right\}$

Question

We say that a sequence $\left\{y_n\right\}$ is a rearrangement of a sequence $\left\{x_n\right\}$ if there is a 1-1 correspondence $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $\forall$ $n \in \mathbb{N}$, $y_n = x_{f(n)}$. Suppose $\left\{y_n\right\}$ is a rearrangement of $\left\{x_n\right\}$. Prove that $\left\{y_n\right\} \rightarrow L$ iff $\left\{x_n\right\} \rightarrow L$.

1) How do I prove this ?

Answer 1

Here is a proof using the definition of convergence with $\varepsilon$'s. It's quite straightforward, and relies on the fact that if all but finitely many elements of $(x_n)_n$ are at distance at most $\varepsilon$ from $\ell$, then you have all but finitely many elements of $(y_n)_n$ that are far from $\ell$ -- basically, the same "finitely many" elements. These elements can be spread out a bit more (by $f$), but since there are finitely many they all must be bounded by some $N$.

1. It is sufficient to prove a single direction (convergence of $(x_n)_n$ implies convergence of $(y_n)_n$), since by symmetry this implies both directions. (Indeed, if $(y_n)_n$ is a rearrangement of $(x_n)_n$, then $(x_n)_n$ is a rearrangement of $(y_n)_n$.)

2. Suppose $(y_n)_n$ is a rearrangement of $(x_n)_n$, and let $f\colon\mathbb{N}\to\mathbb{N}$ be the corresponding bijection. Suppose $(x_n)_n$ converges to some $\ell\in\mathbb{R}$.

   Fix any $\varepsilon > 0$. By assumption, there exists $n_\varepsilon\geq 1$ such that, for every $n\geq n_\varepsilon$, $\lvert x_n - \ell\rvert \leq \varepsilon$. Now consider $$S_\varepsilon = \{ f^{-1}(n) : n < n_\varepsilon \} = f(\{1,2,\dots,n_\varepsilon-1\})\,.\tag{1}$$ This set is finite (it has size $n_\varepsilon-1$), and therefore it has an upper bound $N_\varepsilon$: $$ \forall n \in S_\varepsilon,\qquad n < N_\varepsilon\,. \tag{2} $$ I claim that for every $n\geq N_\varepsilon$ we have $f(n) \geq n_\varepsilon$. (Indeed, suppose by contradiction this is not the case, i.e., there exists $n^\ast \geq N_\varepsilon$ with $f(n^\ast) < n_\varepsilon$. Then $n^\ast = f^{-1}(f(n^\ast)) \in S_\varepsilon$, contradicting $(2)$).

   Therefore, for every $n\geq N_\varepsilon$, we have $f(n) \geq n_\varepsilon$ and therefore $$ \lvert y_n - \ell\rvert = \lvert x_{f(n)} - \ell\rvert \leq \varepsilon\,.\tag{3} $$

   Since $\varepsilon$ was arbitrary, we showed that $(y_n)_n$ converges to $\ell$.