R with the cocountable topology is not first countableby mrbohn1 Tags: cocountable, countable, topology 

#1
Oct2008, 05:21 PM

P: 97

1. The problem statement, all variables and given/known data
(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. 2. Relevant equations (The cocountable topology on R has as its closed sets all the finite and countable subsets of R). 3. 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! 



#2
Oct2008, 06:30 PM

P: 337

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.




#3
Oct2108, 01:26 PM

P: 97

(from wikpedia): "a space, X, is said to be firstcountable 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."




#4
Oct2108, 03:48 PM

Sci Advisor
HW Helper
P: 2,020

R with the cocountable topology is not first countable
Suppose there's a countable basis at 0, say {U_1, U_2, ...}. What is the intersection of all the U_i?



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