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 { } be an infinite sequence in K. By Heine Borel them, K is bounded and closed. By Bolzano-Weistrass thm, { } has a convergent subsequence. Say { }, with lim { } as a -> = . Note that is a limit point of K so since K is closed 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 { } K such that | |->+ as n-> | |>n for all n=1,2,.... Clearly { } has no convert subsequence which is a contradiction to what we proved earlier. If K is not closed, then there is { } K such that lim as n-> = R but K. since K is sequentially compact , there exists a subsequence { } of { } such that lim = as a-> with K. Also lim = since -> . Therefore = , but K and .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.