# 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 { $x_n$} 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.

3. Originally Posted by maddas
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.
hm yours is so much more efficient lol! but im just glad im figuring out proofs. However are you just basically using the definition of sequentially compact when you show that there is a divergent sequence that has a convergent subseqence? I guess that would be the "sequential characterization of compactness"..But when you say lim f(x) is finite, it does make it clear for the next part (which i forgot to add i'm sorry, I took this part from another post bc my origal one was getting crowded and confusing so i just posted the part i was having trouble with) but the next part of the question is : Is the same conclusion valid if we drop the assumption that the limit of f is finite?

4. No, take K=[-1,1] and f(x)=1/x (define it however you want at 0). Then K is compact but f is not bounded.

5. Awesome! thanks so much you've been a BIG help! I couldn't even find the definition in my book for seq. compactness but its in my notes, I was so confused but now I totally get it! thanks for the help

6. Heh, I had to look it up on wikipedia ;]