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

  • Thread starter Thread starter jmjlt88
  • Start date Start date
  • Tags Tags
    Proof Space
Click For Summary
SUMMARY

The discussion focuses on the proof that the set ℝ with the lower-limit topology is Lindelof, as established by Munkres. The Lindelof condition is confirmed to be equivalent to the requirement that every open covering of the space by basis elements has a countable subcollection that covers the space. The construction of the collection B, consisting of basis elements that lie entirely within the open sets of a covering A, is deemed clear and legally valid for completing the proof.

PREREQUISITES
  • Understanding of topology, specifically the lower-limit topology
  • Familiarity with the Lindelof property in topological spaces
  • Knowledge of basis elements in topology
  • Experience with mathematical proofs and constructions in set theory
NEXT STEPS
  • Study the Lindelof property in various topological spaces
  • Explore the implications of basis elements in topology
  • Learn about different types of topologies, including lower-limit topology
  • Review Munkres' "Topology" for detailed explanations and examples
USEFUL FOR

Mathematicians, particularly those specializing in topology, educators teaching advanced mathematics, and students preparing for higher-level proofs in set theory and topology.

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 definition 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! :)
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
8K
Replies
4
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
20
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 45 ·
2
Replies
45
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K