there is one answer viewtopic.php?f=1&p=287#p287

but i don't think it is right:

"for (2), K is bounded is obvious ( let f: K------->K with f(x)=x, them K is bounded)

Now prove that K is closed. Suppose that K is not closed, then R^n\K is not open : since K is bounded so K must have form :

K=(a1,b1)x(a2,b2]x....x(an,bn)

so let F : K--------> R F(x1,x2,....,xn)=[1/(x1-a)]x2....xn then f is continuous and f is not bounded - contradiction

so K <R^n is closed and bounded then K is compact"

Why a bounded but not open set can have the form (a1,b1)x(a2,b2]x....x(an,bn)?

I think open set is really different from open interval,although we have the open set construct theorem,ie every open set are the combination of some countable open intervals. But it is conuntable,not n(finite).

Is there any other proof?

thank you ~~