The image of a compact set under a sequentially continuous real function is bounded

Question

Let $S$ be compact and let $f:S\longrightarrow \mathbf{R}$ be sequentially continuous. Then the image set $f(S)$ is bounded.

Answer 1

Hint: Consider a sequence $x_n$ such as $|f(x_n)|\to \sup_S |f| $.

detail:

There is a subsequence $y_n$ convergent to a certain $y\in S$. $$ \sup_S |f| = \lim |f(x_n)| = \lim |f(y_n)| = |f(\lim y_n)| = |f(y)|<\infty$$

because $|f|$ is sequentially continuous.

Answer 2

The compact Hausdorff space $\beta\mathbb{N}$ has the property that all of its convergent sequences are eventually constant. See this question for an argument.

So any unbounded function $f$ from $\beta\mathbb{N}$ to $\mathbb{R}$ is a counterexample.

Note that if $S$ were sequentially compact (which $\beta\mathbb{N}$ is not), then $f[S]$ would be sequentially compact in $\mathbb{R}$ (if $y_n = f(x_n)$ defines a sequence in $f[S]$, there is a convergent subsequence $x_{n_k}$ converging to some $p \in S$. But then sequential continuity gives that $y_{n_k} = f(x_{n_k}) \rightarrow f(p)$, as $k \rightarrow \infty$.) and in a metric space sequential compactness is equivalent to compactness and so $f[S]$ would be (closed and) bounded.