# Showing the equivalence of two statements regarding the limits of sequences

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?

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 $$L$$.

• 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? Nov 26, 2020 at 13:22
• @user666150: Oops, that was a typo, it should be $|f(a_{n_k}) - L | \ge \epsilon$. Nov 26, 2020 at 13:23
• Thanks, I think I understand it now. Nov 26, 2020 at 13:24