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R with the cocountable topology is not first countable

by mrbohn1
Tags: cocountable, countable, topology
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mrbohn1
#1
Oct20-08, 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!
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jhicks
#2
Oct20-08, 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.
mrbohn1
#3
Oct21-08, 01:26 PM
P: 97
(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."

morphism
#4
Oct21-08, 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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