R with the cocountable topology is not first countable

  • Thread starter Thread starter mrbohn1
  • Start date Start date
  • Tags Tags
    Topology
Click For Summary
SUMMARY

The discussion centers on proving that the real numbers R, equipped with the cocountable topology, are not first countable. The cocountable topology defines closed sets as all finite and countable subsets of R. A key conclusion is that if a point z is in the closure of a subset A, then in a first countable space, there must exist a sequence in A converging to z. The participants highlight the challenge of demonstrating this property through examples and definitions related to first countability.

PREREQUISITES
  • Understanding of cocountable topology and its properties
  • Knowledge of first countable spaces and their definitions
  • Familiarity with concepts of closure and convergence in topology
  • Basic understanding of sequences and neighborhoods in metric spaces
NEXT STEPS
  • Study the properties of cocountable topology in detail
  • Research examples of first countable and non-first countable spaces
  • Explore the implications of closure and convergence in different topological spaces
  • Investigate the relationship between countability and neighborhood bases in topology
USEFUL FOR

Students of topology, mathematicians exploring properties of different topological spaces, and anyone interested in understanding the nuances of first countability in the context of cocountable topology.

mrbohn1
Messages
95
Reaction score
0

Homework Statement



(a) Prove that R, with the cocountable topology, is not first countable.

(b) Find a subset A of R (with the cocountable topology), and a point z in the closure of A such that no sequence in A converges to z.

Homework Equations



(The cocountable topology on R has as its closed sets all the finite and countable subsets of R).

The Attempt at a Solution



I'm not really sure how to prove this. I know that if I could find a set and point meeting the conditions in part (b), then that would prove part (a), as in a first countable space X a point x belongs to the closure of a subset A of that space if and only if there is a sequence of points of A converging to x. However, I assume that I am expected to prove that R is not first countable with this topology some other way for part (a).

Either way, I'm stuck!
 
Last edited:
Physics news on Phys.org
When you say *first* countable do you mean isomorphic to Z? Maybe it would be easier to show that the cocountable topology on Z isn't countable.
 
(from wikpedia): "a space, X, is said to be first-countable if each point has a countable neighbourhood basis (local base). That is, for each point, x, in space X there exists a sequence, U1, U2, … of open neighborhoods of x such that for any open neighborhood, V, of x, there exists an integer, i, with Ui contained in V."
 
Suppose there's a countable basis at 0, say {U_1, U_2, ...}. What is the intersection of all the U_i?
 

Similar threads

Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K