Sequences proof exercise

Question

I was asked this in an exam recently and couldn't get to an answer: Let $\;\mathit f:\Bbb R \to \Bbb R$ be a continuous function and $(a_n)_{n\in\Bbb N} \subset \Bbb R$ a bounded sequence. Prove: $$\; \mathit f(PL\{(a_n)_{n\in\Bbb N}\} = PL \{\mathit f(a_n)_{n\in\Bbb N}\}$$ where PL is the set of all limit points of $a_n$.

I can prove that $\mathit f(PL\{(a_n)_{n\in\Bbb N}\} \subset PL \{\mathit f(a_n)_{n\in\Bbb N}\}$, but I can't prove the other way.

Answer 1

Take a sub-sequence $\{a_{n(i)}\}_{i\in\mathbb N}$ and some $a\in\mathbb R$ such that $(f(a_{n(i)}))\rightarrow a$ (i.e. $a$ is the limit point from a convergent subsequence of $f(a_n)_{n\in\mathbb N}$, or in the notation you used $a\in PL\{f(a_n)_{n\in\mathbb N}\}$). You want to prove there is some $x\in PL\{(a_n)_{n\in\mathbb N})\}$ such that $f(x) = a$.

Since $(a_{n(i)})_{i\in\mathbb N}$ is bounded, it has converging sub-sequence, for simplicity, lets denote that as $(a_{k(j)})_{j\in\mathbb N}$ and let $x$ be the limit point from this sub-sequence.

Therefore, $f(a_{k(j)})_{j\in\mathbb N}$ is a sub-sequence from $f(a_{n(i)})_{i\in\mathbb N}$. Thus, $f(a_{k(j)})_{j\in\mathbb N}$ converges to $a$ (every sub-sequence from a converging sequence is converging and converges to the same point as the original sequence). However, since $f$ is continuous, then $f(a_{k(j)})_{j\in\mathbb N}$ converges to $f(x)$. Therefore, $f(x)=a$ and $x\in PL\{(a_n)_{n\in\mathbb N}\}$.

Answer 2

If $x$ lies in $PL\{f(a_n)_{n\in\mathbb{N}}\}$, then there is some subsequence $(f(a_{n_k}))_{k\in\mathbb{N}}$ of $(f(a_n))_{n\in\mathbb{N}}$ tending to $x$. Now, $(a_{n_k})_{k\in\mathbb{N}}$ is a bounded sequence (since $(a_n)_{n\in\mathbb{N}}$ is), so it has a convergent subsequence $\left(a_{n_{k_m}}\right)$. Let $y$ be the limit of $\left(a_{n_{k_m}}\right)$. Then, since $f$ is continuous, $f(y) = f\left(\lim\limits_{m\to\infty} a_{n_{k_{m}}}\right) = \lim\limits_{m\to\infty} f\left(a_{n_{k_m}}\right) = \lim\limits_{k\to\infty} f(a_{n_k}) = x$, so $x = f(y)$, and since $y$ lies in $PL\{(a_n)_{n\in\mathbb{N}}\}$, we have the result.