Let , then prove that:
is compact is closed and bounded.
is compact then is bounded:
The open intervals have union , so they definitely cover . since is compact and the above covering is open so it admits a finite subcovering which means for some and hence is bounded.
now to show that A is closed too.
here is the part where i am stuck:
Let be a sequence in such that . I must show that . i tried to use contradiction method, that is, i assumed that there exists a sequence in with such that . now i must exhibit an open covering of which does NOT admit a finite subcovering in order to bring up a contradiction but i am not able to go further.