In my general topology textbook there is the following exercise:
A topological space $(X,\tau)$ is said to satisfy the second axiom of countability if there exists a basis $B$ for $\tau$, where $B$ consists of only a countable number of sets.
- Let $(X,\tau)$ be the set of all integers with the finite-closed topology. Does the space $(X,\tau)$ satisfies the second axiom of countability.
I made a proof for this but in my prove I made a mistake that I'll point out with the number (1).
My proof
Let $A_i=\{\mathbb Z\setminus B : \text{card}\ B=i\}$, for $B \subset \mathbb Z$. Then we have that $\tau= \bigcup \limits _{i=0} ^\infty A_i$. If every set $A_i$ is countable then we have that $\tau$ is countable as well.
Let's fix a value of $i$ and define the set $C_i=\{X: \text{card} \ X=i\}$, for $X \subset \mathbb Z$.
Each $X$ is countable because it's finite, and because $C_i=\bigcup X$ (1), then $C_i$ is countable.
Let $f_i: C_i \to A_i$, with $f_i(X)=Z\setminus X$, then $f_i$ is bijective, thus $C_i \sim A_i$, thus $A_i$ is countable, proving that $\tau$ is also countable.
If $B$ is a basis for this space, then $B \subset \tau$, any subset of a countable set is also countable thus $B$ is countable, so $(\mathbb Z, \tau)$ does satisfy the axiom.
My mistake was that $C_i \neq \bigcup X$, but instead $C_i = \bigcup \{X\}$. How can I prove that $C_i$ is countable thus concluding the proof?
Best Answer
You are right that $C_i\neq \bigcup X$ and that rather $C_i \bigcup {X}$. I think it is easiest to prove $C_i$ is countable by induction on $i$. For $i=0$ this is trivial (in fact $C_0$ has only one element: the empty set), and it's still pretty trivial for $i=1$ (since $C_1$ is more or less $\mathbb{Z}$ itself). Now, supposing $C_n$ is countable, it is pretty easy to show $C_{n+1}$ is countable as well.
For instance, you can say that, since $C_n$ is countable and $\mathbb{Z}$ is countable, also $C_n\times \mathbb{Z}$ is countable, and hence $S:=\{(A,z):A\in C_n,z\in\mathbb{Z}\setminus A\}$ is countable, and we a surjective map from $S$ onto $C_{n+1}$ (sending $(A,z)$ to $A\cup \{z\}$.
Thus, $C_n$ is countable for every $n$.
By the way, a remark: you proved that every basis of this space is countable. This is correct, but you didn't need to show that. You only needed to show that there exists a countable basis of this space.