# R with the cocountable topology is not first countable

1. Oct 20, 2008

### mrbohn1

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!

Last edited: Oct 20, 2008
2. Oct 20, 2008

### jhicks

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. Oct 21, 2008

### mrbohn1

(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."

4. Oct 21, 2008

### morphism

Suppose there's a countable basis at 0, say {U_1, U_2, ...}. What is the intersection of all the U_i?