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Given a sequence $\{y_k\}$ is obtained by permuting the sequence $\{x_n\}$ and: $$ \{\forall n\in \mathbb N\ \exists k_n \in \mathbb N:x_n = y_{k_n}, n_1 \ne n_2 \implies k_{n_1} \ne k_{n_2}\} $$ and: $$ \{\forall k\in \mathbb N\ \exists n_k \in \mathbb N:y_k = x_{n_k}, k_1 \ne k_2 \implies n_{k_1} \ne n_{k_2}\} $$ Prove that: $$ \lim_{n\to \infty}x_n = a \implies \lim_{k\to \infty}y_k=a $$

I've been thinking of the following. Let there exist a bijection $P:\mathbb N\to \mathbb N$ which converts $k$ to $n_k$. So basically we have that if $x_n$ is convergent to some $a$ then:

$$ |x_n - a| < \varepsilon $$

Thus : $$ \forall \varepsilon>0 \ \exists N \in \mathbb N:n \ge N \implies |x_n - a| < \varepsilon $$

This is only valid after some $N$. The only terms violating that condition are $x_1, x_2, \dots, x_N$. My guess was that we could take $\max\{P(n)\}$ and that would imply that $|y_{P(n)} - a| < \varepsilon$, so: $$ \forall \varepsilon>0 \ \exists\ M = \max\{n: 1 \le P(n) \le N\}: n>M \implies |y_n-a| < \varepsilon $$

I decided to test this idea graphically and here is what I got. $$ x_n = \frac{10}{n} $$

So from the graph for $\varepsilon = 4$ it follows that for $N \ge 3 $ all terms of $x_n$ are satisfying the condition of convergence. Now if we take the maximum value for $P(n)$ where $n \le 3$ we get $16$. But after $y_{16}$ there exists a value $y_{19}$ which falls out of the neighborhood of $0$.

Where did I get it wrong and how to prove the statement in the problem section?

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if $\{a\}$ converges then for any $\epsilon > 0$ there is an $N>0$ such than $n>N \implies |a_n - a| < \epsilon$

This means that there are only finitely many $a_n$ such that $|a_n - a| > \epsilon$

When we permute $\{a\}$ to construct $\{b\}$ then there are only finitely many $b_m$ such that $|b_m - a| > \epsilon$

And we can choose $M$ such that $M$ is greater than the largest index associated with $|b_m - a| < \epsilon$

and $m>M \implies |b_m - a| < \epsilon$

What is the problem with your graph? In your graph, you have only plotted finitely many members of the sequence.

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