If $f(a_{n_k}) \to a$ as $k \to \infty$, then there is some $x \in \mathbb{R}$ such that $f(x) = a$ and $a_{n_k} \to x$.

Question

I have two questions:

Let $(a_n)_{n \in \mathbb{N}}$ be a bounded sequence and let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function.

If $a$ is an limit point for $f(a_n)_{n \in \mathbb{N}},$ there is there some $y \in \mathbb{R}$ such that $f(y) = a$ and $y$ is an accumulation point for $(a_n)$?

And also:

If $a$ is an limit point for $f(a_n)_{n \in \mathbb{N}},$ and there is some $y \in \mathbb{R}$ such that $f(y) = a$, then $y$ is an accumulation point for $(a_n)$.

I have tried writing the definition of continuity with $\varepsilon - \delta$ and with sequences, but I did not manage to solve the above problems, because that definition gives no information on $(a_n)$.

I have also tried constructing a counter example for the first question (since I believe to be false), but I was not able to.

Thank you in advance for the help!

Answer 1

If $a$ is a limit point of $\{f(a_n)\}$, then there exists a subsequence $a_{k_n}$, such that $$ f(a_{k_n})\to a $$ and since $\{a_{n_k}\}$ is bounded, then it possesses a convergent subsequence, say $a_{\ell_n}\to b$. Now we have that $$ a_{\ell_n}\to b\qquad\text{and}\qquad f(a_{\ell_n})\to f(b) $$ and also $f(a_{\ell_n})\to a$.

Thus $f(b)=a$.