Register to reply 
Showing a closed subspace of a Lindelöf space is Lindelöf 
Share this thread: 
#1
Nov1710, 12:47 PM

HW Helper
P: 3,224

1. The problem statement, all variables and given/known data
As the title says, one needs to show that if A is a closed subspace of a Lindelöf space X, then A is itself Lindelöf. 3. The attempt at a solution Let U be an open covering for the subspace A. (An open covering for a set S is a collection of open sets whose union equals S, btw some define a cover for a set S as a collection such that S is contained in the union of these sets, it seems this causes disambiguity sometimes?) Since all elements of U are open in A, they equal the intersection of some family of open sets with A, call it U'. Now, consider the open cover for X consisting of X\A and U'. By hypothesis, this cover has a countable subcover. Dismiss X\A from it, and then the intersection of the elements left in this collection with A form a countable open cover for A. I hope this works. 


#2
Nov1710, 02:22 PM

Mentor
P: 18,036

This is 100% correct!
I'm aware that the notion of "cover" is sometimes defined in another way. But this never causes any problems. The two notions are interchangeable. 


Register to reply 
Related Discussions  
Subspace of l2/L2 that is closed/not closed.  Calculus & Beyond Homework  1  
Showing a set is closed with the definition of continuity  Calculus & Beyond Homework  3  
Showing a graph is closed  Differential Geometry  4  
Not closed linear subspace  General Math  7  
Two closed subspace whose sum is not closed?  Calculus  3 