Suppose $f:K \to (-\infty, \infty), 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. Then what if we don't know if $f$ is finite.
I'm really confused on this one. Can anyone give me some guidance?
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Please explain: "Then what if we don't know if $f$ is finite?" But you described the function as mapping $K$ into $(-\infty,\infty)$.
Here is what you know:
At every point $x\in K$ you know there is a $\delta(x)>0$ so that $f$ is bounded on both $(x-\delta(x),x)\cap K $ and $(x,x+\delta) \cap K$. But $f(x)$ is finite so in fact $f$ is bounded on $(x-\delta(x),x+\delta(x))\cap K $. That is the situation you are in and you should be able to construct a proof.
When you are doing this kind of compactness problem you should try to be aware of just how similar they all are. I would describe it this way: if there is a local condition [state condition] that holds at every point of a compact set $K$, then that condition holds globally on $K$.
You must have seen this many times: continuous locally at each point of a compact set implies uniform continuity. Or, here locally bounded implies globally bounded. Or locally constant on $[a,b]$ implies constant. Locally increasing on $[a,b]$ implies increasing. Locally Lipschitz on $[a,b]$ implies globally Lipschitz. Etc.
Since $f$ has a finite limit at each point of $K$, for each point $x_0\in K$, there is a $\delta>0$ such that $|f(x)-L|<1$, i.e. $f$ is bounded in $(x_0-\delta, x_0+\delta)$. Since it holds for any point $x_0\in K$, $K\subset\bigcup_{x_0\in K}(x_0-\delta, x_0+\delta)$, i.e. an open cover of $K$. Since $K$ is compact, there is a finite subcover that $K\subset\bigcup_{1\leqslant i\leqslant N}(x_i-\delta, x_i+\delta)$. Then $|f|<M$ on $K$ if we set $M=\max{(M_1,\cdots, M_N)}$, where $|f|<M_i$ on $(x_i-\delta, x_i+\delta)$. So $f$ is bounded on $K$.
If $f$ doesn't have finite limit at each point of $K$, $f$ may not be bounded on $K$ for it may not be bounded on $(x_0-\delta, x_0+\delta)$.