Suppose f: K->(-,), K is compact, and f has a finite limit at each point of K, but may not be continuous on K. Show that f is bounded by using the sequential characterization of compactness.

Would showing the proof of the fact that K is compact iff K is sequentially compact work for showing that f is bounded?

as in can i do this:

Assume K is compact and show K is sequentially compact. Let {$\displaystyle x_n$} be an infinite sequence in K. By Heine Borel them, K is bounded and closed. By Bolzano-Weistrass thm, {$\displaystyle x_n$} has a convergent subsequence. Say {$\displaystyle x_n$$\displaystyle _a$}, with lim {$\displaystyle n_a$} as a -> $\displaystyle \infty$ = $\displaystyle \bar{x}$. Note that $\displaystyle \bar{x}$ is a limit point of K so since K is closed $\displaystyle \bar{x}$$\displaystyle \in$K.

Now assume K is sequentially compact and show K is compact so suppose not. Then either K is not bounded or K is not closed. If K is not bounded, there exists {$\displaystyle x_n$}$\displaystyle \subset$K such that |$\displaystyle x_n$|->+$\displaystyle \infty$ as n->$\displaystyle \infty$ |$\displaystyle x_n$|>n for all n=1,2,.... Clearly {$\displaystyle x_n$} has no convert subsequence which is a contradiction to what we proved earlier. If K is not closed, then there is {$\displaystyle x_n$}$\displaystyle \subset$K such that lim$\displaystyle x_n$ as n->$\displaystyle \infty$= $\displaystyle \bar{x}$$\displaystyle \in$R but $\displaystyle \bar{x}$$\displaystyle \notin$K. since K is sequentially compact , there exists a subsequence {$\displaystyle x_n$$\displaystyle _a$} of {$\displaystyle x_n$} such that lim$\displaystyle x_n$$\displaystyle _a$=$\displaystyle \bar{y}$ as a-> $\displaystyle \infty$ with $\displaystyle \bar{y}$$\displaystyle \in$K. Also lim$\displaystyle x_n$$\displaystyle _a$ = $\displaystyle \bar{x}$ since $\displaystyle x_n$->$\displaystyle \bar{x}$. Therefore $\displaystyle \bar{x}$=$\displaystyle \bar{y}$, but $\displaystyle \bar{x}$$\displaystyle \not in$K and $\displaystyle \bar{y}$$\displaystyle \in$.Thus a contradiction. therefore K must be closed and BOUNDED.

I'm not sure, I seem to be stuck on this problem. I don't know if its appropriate to show the whole proof of sequentially compact or if im doing this wrong or what.