Question:

Let , then prove that:

is compact is closed and bounded.

approach:

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.