Claim made in a proof that a particular space is Lindelof.

  • Thread starter Thread starter jmjlt88
  • Start date Start date
  • Tags Tags
    Proof Space
jmjlt88
Messages
94
Reaction score
0
I was reading over the proof that the set ℝ with the lower-limit topology is Lindelof. Munkres claimed the Lindelof condition is equivalent to the condition that every open covering of the space by basis elements has a countable subcollection that cover the space. I wanted to write out all the details of the verification of this claim. Basis elements are open sets; hence, one direction is easy. For the other direction, we assume the space X satisfies the condition involving its basis elements and let A be an covering of X by sets opem in X. For each x in X, there is some element of A that contains x; denote it Ax. By the defintion of a basis for a topology, there is some Bx that contains x and lies entirely in Ax.
-----The Question-----
Now, I want to let B be the collection of these basis elements. That is, B is the collection of all basis elements Bx such that Bx contains x and lies entirely in some element Ax of A containing x. Is this construction clear (and legal)? Given an arbitrary basis element B, if for one of its elements x, B lies entirely in some element of A contaning x, then B is a member of B. If this collection is well-defined, then I can finish off the details.

Thank you!
 
Physics news on Phys.org
Sure, that's the way to do it.
 
Thanks micromass! :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top