Vtyjkyuk

We say that a sequence $lefty_nright$ is a rearrangement of a sequence $leftx_nright$ if there is a 1-1 correspondence $f : mathbbN rightarrow mathbbN$ such that $forall$ $n in mathbbN$, $y_n = x_f(n)$. Suppose $lefty_nright$ is a rearrangement of $leftx_nright$. Prove that $lefty_nright rightarrow L$ iff $leftx_nright rightarrow L$.

1) How do I prove this ?

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 $fcolonmathbbNtomathbbN$ be the corresponding bijection. Suppose $(x_n)_n$ converges to some $ellinmathbbR$.

   
   
   

   Fix any $varepsilon > 0$. By assumption, there exists $n_varepsilongeq 1$ such that, for every $ngeq n_varepsilon$, $lvert x_n - ellrvert leq varepsilon$.
   Now consider $$S_varepsilon = f^-1(n) : n < n_varepsilon = f(1,2,dots,n_varepsilon-1),.tag1$$
   This set is finite (it has size $n_varepsilon-1$), and therefore it has an upper bound $N_varepsilon$:
   $$
   forall n in S_varepsilon,qquad n < N_varepsilon,. tag2
   $$
   I claim that for every $ngeq N_varepsilon$ 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 $ngeq N_varepsilon$, we have $f(n) geq n_varepsilon$ and therefore
   $$
   lvert y_n - ellrvert = lvert x_f(n) - ellrvert leq varepsilon,.tag3
   $$

   
   
   

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