If $X$ is the set of positive integers, how can some topology on it not satisfy the first axiom of countability?

Question

Let $X$ be the set of integers. Obviously, $X$ satisfies the second axiom of countability, which implies it also implies the first axiom of countability. Regardless of the topology!

Let $X$ be Appert's space (http://en.wikipedia.org/wiki/Appert_topology). My textbook says this space does not satisfy the first axiom of countability. How this is possible?

Thank you for your time.

$X$ contains a countable number of elements- the set of positive integers. The number of base sets also have to be countable. I think....? — , Jun 12 '13 at 17:54
My logic was: the cardinality of $\mathbb{N}$ is $\mathfrak{N_{0}}$. The number of possible base sets, if every possible combination of positive integers is considered, is $2^{\mathfrak{N_{0}}}$. Isn't this again a countable number? — , Jun 12 '13 at 17:58
Another example of a countable space which is not first countable is the Arens–Fort space, which I describe a bit more fully in this previous answer. — user642796, Jun 12 '13 at 20:18

score 3 · Answer 1 · answered Jun 12 '13 at 19:54

3

The countability axioms refer to the cardinality of a base, not the cardinality of the ground set. As you mention in the comments, a topology on a countable set may require up to $2^{\aleph_0}$ open sets for its base, which is uncountable.

answered Jun 12 '13 at 19:54

Austin Mohr

26,272

score 0 · Answer 2 · answered Jun 13 '13 at 00:45

First let us recall the definiton of Apperts topology.

Let $X$ be the set of positive integers. Let $N(n,E)$ denote the number of integers in a set $E\subset X$ which are less than or equal to $n$. We describe Apperts topology on $X$ by declaring open any set which excludes the integer 1, or any set $E$ containing 1 for which $\lim_{n\to \infty} \frac{N(n,E)}{n}=1$.

Now we will pove that it is not first countable.

Proof: $X$ is not first countable, since the point 1 hasn't a countable local base at 1. Suppose that $\{B_n\}_n$ were a countable base at 1. Then each $B_n$ must be infinite, so we can select a point $x_n\in B_n$ such that $x_n>10^n$; then $U_n=X\setminus \{x_n\}_n$ does not contain any of the sets $B_n$, yet it is an open nbhd of 1, for $\lim_{n\to \infty} \frac{N(n,U)}{n}=1$.

If $X$ is the set of positive integers, how can some topology on it not satisfy the first axiom of countability?

2 Answers2