2
$\begingroup$

I need to show the equivalence of the following two statements

$(i)$ For any sequence $(a_n)_n$ with $a_n \in D$ and $a_n > a$ for all $n \in N$ such that $\lim\limits_{n\to\infty} a_n = a$ we have $\lim\limits_{n\to\infty} f(a_n) = L.$

$(ii)$ For any sequence $(a_n)_n$ with $a_n \in D$ and $a_n \ge a_{n+1}>a$ for all $n \in N$ such that $\lim\limits_{n\to\infty} a_n = a$ we have $\lim\limits_{n\to\infty} f(a_n) = L.$

My attempt:

Going from $(i)$ to $(ii)$ is obvious.

To show that $(ii)=>(i)$, take a sequence $(a_n)_n$ with $a_n \in D$ and $a_n > a$ for all $n \in N$ and $\lim\limits_{n\to\infty} a_n = > a$. We want to show that $\lim\limits_{n\to\infty} f(a_n) = L$.

$(a_n)_n$ has a non-increasing subsequence $(a_{k_n})_n$. Then, we have $a_{k_n} \in D$ and $a_{k_n} \ge a_{k_{n+1}}>a$ for all $n \in N$ and $\lim\limits_{n\to\infty} a_{k_n} = a$. Then, we can apply $(ii)$ to this sequence and get $\lim\limits_{n\to\infty} f(a_{k_n}) = L$.

I am stuck here and I don't know how to go back to the limit of the original sequence. Am I on the right path? Can I get some hints?

$\endgroup$

1 Answer 1

1
$\begingroup$

Take a sequence $(a_n)_n$ with $a_n \in D$ and $a_n > a$ for all $n \in N$ and $\lim_{n\to\infty} a_n = a$.

Assume that $\lim_{n\to\infty} f(a_n) = L$ does not hold. Then there is a $\epsilon > 0$ and a subsequence $(a_{n_k})$ of $(a_n)$ such that $|f(a_{n_k}) - L | \ge \epsilon$ for all $k$.

Now apply (ii) to a decreasing subsequence of $(a_{n_k})$ to get a contradiction.

This is an example of the following principle, applied to $x_n = f(a_n)$:

Let $(x_n)$ be a sequence in a topological space $X$ and $L \in X$. If every subsequence of $(x_n)$ has itself a subsequence converging to $L$, then the complete sequence $(x_n)$ converges to $L$.

$\endgroup$
3
  • $\begingroup$ In the second paragraph, we assumed that the limit of $f(a_n)$ is not L. Then, how did we get that there is such a subseqeunce and what happened to the function $f$, can you please explain that part? $\endgroup$
    – user666150
    Nov 26, 2020 at 13:22
  • 1
    $\begingroup$ @user666150: Oops, that was a typo, it should be $|f(a_{n_k}) - L | \ge \epsilon$. $\endgroup$
    – Martin R
    Nov 26, 2020 at 13:23
  • $\begingroup$ Thanks, I think I understand it now. $\endgroup$
    – user666150
    Nov 26, 2020 at 13:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .