Assume this is false. Then for we can choose such that . Therfore, is a sequence in - a sequencially kompact set. Thus, we can find a convergent subsequence such that there is (with ) so that . We shall show this leads to a contradiction. Let . Since is closed it follows . Now since is open there is such that . Since it means there is such that if for . Now and eventually gets smaller than . Thus there is a so that .

This is a contradiction.