Question on convergence involving one-to-one functions

Question

Could someone please explain how the following problem is solved?

Suppose that $lim_{n\to\infty}s_n=c$ and that $\sigma:\Bbb{N} \to \Bbb{N}$ be one-to-one. Prove that $lim_{n\to\infty}s_{\sigma(n)}=c$.

Hint : Since $\sigma$ is one-to-one, then the set {$n\in\Bbb{N}$: $ \sigma(n) \leq N $} is finite for any N $\in \Bbb{N}$.

Throw-away attempt :

Assume $<s_n>$ converges to c and let $\sigma:\Bbb{N} \to \Bbb{N}$ be one-to-one. Let $\epsilon \gt 0$ be a real number. The idea,if i understand it corectly, invovlves reverse engineering a natural number $N \in \Bbb{N}$ such that for all $n \gt N$, $|s_{\sigma(n)}-c| \lt \epsilon$.

Therefore:

$|s_{\sigma(n)}-c| = |(s_{\sigma(n)}-s_n)+(s_n-c)| \leq |(s_{\sigma(n)}-s_n)|+|(s_n-c)|$

...now since $<s_n>$ converges to c then by definition there exists some $N_s \in \Bbb{N}$ such that for $n \gt N_s$ , $|s_n-c|$ can be made as small as necessary..

but what do i do with the other term $|(s_{\sigma(n)}-s_n)|$ ? and how is the hint involved ?

Thank you for any clarifications.

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New Attempt :

Suppose,for a contradiction,that $lim_{n\to\infty}s_{\sigma(n)}=d$ where d is not equal to c. By definition, for every $U_\epsilon$ of d, there is a $\sigma(N) \in \Bbb{N}$ s.t. for all $\sigma(n) \gt \sigma(N)$, $s_{\sigma(n)} \in U_{\epsilon}$. By corollary, every $U_{\epsilon}$ contains an infinite number of points from the sequence $<s_{\sigma(n)}>$.

However,setting $N \geq \sigma(n_p)$ where $s_{\sigma(n_p)} \gt d+ \epsilon$ yields two contradictory results :

(i) By the hint that {$n\in\Bbb{N}$: $ \sigma(n) \leq N $} is finite for any N $\in \Bbb{N}$, it follows that there are a finite number of $\sigma(n)$ points.

(ii) Now,if there are infinite $s_{\sigma(n)}$ terms in $U_{\epsilon}$,there must be infinite points of $\sigma(n)$ to serve as the index of the terms of the sequence located in $U_{\epsilon}$.

er....does this make a bit more sense??

Thanks!

Answer 1

This question is very easy if you argue by contradiction, can you see?

Assume your subsequence converges to $d\neq c$.

Then you can show that your original sequence is not convergent, since for any $N$ there always exist infinity many $n$ such that $n>N$, $\sigma(n)$ is closer to $d$ than to $c$.

This statement can be easily made mathematical rigorous using $\epsilon$, care to do the honor?

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I will do the honor:

$\exists N$ such that $\forall n>N$ and $\sigma(n)>N$, $|s_n-c|\leq\frac{|d-c|}{4}$ and $|s_{\sigma(n)}-d|\leq\frac{|d-c|}{4}$.

(infinitely many terms in $\sigma(n)$ satisfy this since by the hint there are only finitely many before $N$)

Hence $|s_n-s_{\sigma(n)}|\geq\frac{|d-c|}{2}$. Since $s_{\sigma(n)}$ is also in $s_n$, the sequence $s_n$ diverges. Contradiction.

(In this last statement I cooked up some distance such that $s_n$ and $s_{\sigma(n)}$ must be that close to their limits. This makes a gap between the terms in them, which creates the contradiction)

Answer 2

We have $\lim_{n\to \infty}s_n=c$ iff for every $r>0$ the set $S(r)=\{n: |s_n-c|\ge r\}$ is finite. (This could even be used as the definition of a limit.)

And $\lim_{n\to \infty} s_{\sigma(n)}=c$ iff for every $r>0$ the set $\Sigma(r)=\{n:|s_{\sigma(n)}-c|\ge r\}$ is finite.

Let $T(r)=\sigma^{-1}S(r)=\{n: \sigma(n)\in S(r)\}.$ Then $T(r)$ is finite because $S(r)$ is finite and $\sigma$ is $1$-to-$1$.

And $T(r)=\Sigma(r).$