# 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

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.

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Yes, exactly, do so. –  Hagen von Eitzen Dec 9 '12 at 16:32
@HagenvonEitzen I have no idea how to go about it though? I don't even understand the hint. –  Mathlete Dec 9 '12 at 16:34

Suppose that $f$ is not bounded above that is $$\forall M\in \mathbb{R} \exists x\in [a,b]:f(x)>M$$ Taking $M=1,...,n,...$ we can construct inductively a sequence $(x_n)\in [a,b]$ such as that $$$$\forall n\in \mathbb{N}\;f(x_n)>n$$$$ The sequence $(x_n)$ is bounded and so by the Bolzano-Weierstrass Theorem it has a convergent subsequence $x_{k_n}\to c$ Since $\forall n\in \mathbb{N}\ a\le x_n \le b$ we have that $a\le c \le b$. By hypothesis, $f$ is sequentially continuous at $c$ and because $x_{k_n}\to c$ it follows $$f(x_{k_n})\to f(c)$$ which is a contradiction to the fact that $$$$\forall n\in \mathbb{N}\;f(x_n)>n$$$$ Thus, $f$ is bounded above.
EDIT: We have $\forall n\in \mathbb{N}\;f(x_n)>n$. This implies $$\lim_{n\to \infty}f(x_n)=+\infty$$ which is a contradiction to the fact that $f(x_{k_n})\to f(c)$
It is bounded above by say, $b$. –  Nameless Dec 9 '12 at 16:52
Take the limit of the inequality as $n\to +\infty$ –  Nameless Dec 9 '12 at 16:54