Let $f$ be a real valued sequentially continuous function relative to a closed bounded interval $I=[a,b]$. Prove that the set $f(I)$ is bounded above

Question

The hint that I've been given is: for each n in the naturals, use the assumption that $n$ is not an upper bound for $f(I)$ to choose a sequence of $x_n$ (from $n=1$ to infinity) in $I$; then apply Bolzano-Weierstrass.

Answer 1

Suppose that $f$ is not bounded above that is \begin{equation}\forall M\in \mathbb{R} \exists x\in [a,b]:f(x)>M\end{equation} Taking $M=1,...,n,...$ we can construct inductively a sequence $(x_n)\in [a,b]$ such as that \begin{equation}\forall n\in \mathbb{N}\;f(x_n)>n\end{equation} The sequence $(x_n)$ is bounded and so by the Bolzano-Weierstrass Theorem it has a convergent subsequence $x_{k_n}\to c$ Since $\forall n\in \mathbb{N}\ a\le x_n \le b$ we have that $a\le c \le b$. By hypothesis, $f$ is sequentially continuous at $c$ and because $x_{k_n}\to c$ it follows \begin{equation}f(x_{k_n})\to f(c)\end{equation} which is a contradiction to the fact that \begin{equation}\forall n\in \mathbb{N}\;f(x_n)>n\end{equation} Thus, $f$ is bounded above.

EDIT: We have $\forall n\in \mathbb{N}\;f(x_n)>n$. This implies \begin{equation}\lim_{n\to \infty}f(x_n)=+\infty\end{equation} which is a contradiction to the fact that $f(x_{k_n})\to f(c)$