# Math Help - sequentially compact? Show bounded?

1. ## sequentially compact? Show bounded?

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

2. Its certainly a valid way to prove it, even if its probably not what the intent of the question was. I like it, in any case :]

If you want another way: suppose f is not bounded. Then chuse $x_n$ s.t. $f(x_n)$ diverges to infinity. But there is a convergent subsequence $x_{k_n}$ which converges to y and $\lim_{x\to y}f(x)$ is finite, a contradiction.

If you want another way: suppose f is not bounded. Then chuse $x_n$ s.t. $f(x_n)$ diverges to infinity. But there is a convergent subsequence $x_{k_n}$ which converges to y and $\lim_{x\to y}f(x)$ is finite, a contradiction.