Does a continuous function map a bounded sequence to a bounded sequence?

Question

Let $f: \mathbb R \to \mathbb R$ be continuous function. Does $f$ map every bounded sequence to a bounded sequence?

Since we are talking about bounded sequence, so maybe we can take $\{1/n\}$ in $(0,1)$. It is bounded here. If we take continuous function $x\mapsto \dfrac 1x$, it would map the sequence to $\{n\}$, which is unbounded.

So I guess the answer to the given question should be No. But answer key says Yes.

Answer 1

Let $(a_n)$ be a bounded sequence. Then there exists a closed interval $[p,q]$ such that $a_n\in[p,q]$, for all $n$.

Since $f$ is continuous, the image of the compact set $[p,q]$ is a closed and bounded set (and also an interval, actually).

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Why does this fail for the sequence $(1/n)$ and $f(x)=1/x$, over $\mathbb{R}_{>0}$? There is no closed and bounded interval $[p,q]$ cointained in $\mathbb{R}_{>0}$ such that $1/n\in[p,q]$, for all $n$.

Answer 2

Indeed, the answer is yes. The map $x \mapsto \frac{1}{x}$ is not continuous on $\mathbb{R}$, it is only continuous over $\mathbb{R}_{-}^* $ or $\mathbb{R}_{+}^*$, this is why close to $0$ a problem arises.

Suppose $(x_n)_{n \in \mathbb{N}}$ is a real bounded sequence and that $f$ is a continuous function over all $\mathbb{R}$. Then, if $M$ is a positive real number such that $ \forall n \in \mathbb{N}, x_n \in [-M,M]$, then, for all positive integers $n$, you must have $f(x_n) \in f\left( [-M,M] \right)$ which is a compact subset of $\mathbb{R}$ since $f$ is continuous. Therefore $\left(f(x_n)\right)$ is bounded.