# Thread: 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 {$\displaystyle x_n$} 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}$$\displaystyle \inR but \displaystyle \bar{x}$$\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}$$\displaystyle \inK. Also lim\displaystyle x_n$$\displaystyle _a$ = $\displaystyle \bar{x}$ since $\displaystyle x_n$->$\displaystyle \bar{x}$. Therefore $\displaystyle \bar{x}$=$\displaystyle \bar{y}$, but $\displaystyle \bar{x}$$\displaystyle \not inK and \displaystyle \bar{y}$$\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.

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 $\displaystyle x_n$ s.t. $\displaystyle f(x_n)$ diverges to infinity. But there is a convergent subsequence $\displaystyle x_{k_n}$ which converges to y and $\displaystyle \lim_{x\to y}f(x)$ is finite, a contradiction.

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