Properties of Infinite set on co-finite topology and Countable set on co-countable topology

Question

I am trying to verify some of the properties of infinite set on co-finite topology and countable set on co-countable topology but it is proven to be very tricky because I cannot visualize the spaces. Can someone verify my solution is correct?

1. Infinite set on co-finite topology: $A = (\mathbb{N}, \mathfrak{T}_{co-finite})$

• Open sets: All co-finite sets

• Closed sets: All finite sets

• Separation Axioms (0-4): $A$ is $T_0$ since $\{x\}$ is contained in $\mathbb{N}\backslash\{y\}$, for $x \neq y$.

  $A$ is $T_1$ since $\{x\}$ is contained in $\mathbb{N}\backslash\{y\}$, $\{y\}$ is contained in $\mathbb{N}\backslash\{x\}$ for $x \neq y$.

  $A$ is not $T_2$ since any open set touches. Therefore not $T_3, T_4$

• Separable: We know that all open set touches in this space, and $\mathbb{N}$ is countable, so there are countably many open sets, so $A$ is separable [something wrong here fixing]

• 2nd Countable (countable base): Not sure...

• 1st Countable (countable local base): Not sure...

2. Countable set on co-countable topology: $B = (\mathbb{N}, \mathfrak{T}_{co-countable})$

• Open sets: All co-countable sets

• Closed sets: All countable sets

• Separation Axioms (0-4): $B$ is $T_0$ since $\{x\}$ is contained in $\mathbb{N}\backslash\{y\}$, for $x \neq y$.

  $B$ is $T_1$ since $\{x\}$ is contained in $\mathbb{N}\backslash\{y\}$, $\{y\}$ is contained in $\mathbb{N}\backslash\{x\}$ for $x \neq y$.

  I am not sure if $B$ is $T_2$ and above. I only know that uncountable set equipped with co-countable topology is not $T_2$

• Separable: Might not be separable, I am not sure how to prove this.

• 2nd Countable (countable base): Not sure...

• 1st Countable (countable local base): Not sure...

Do these spaces even have bases for us to talk about countability?

Answer 1

For (1), you get separable because every countable space is separable (the entire set is a countable dense subset!). It is second countable because the set of finite subsets of a countable space is countable (so there are only countably many closed sets, implying there are also only countably many open sets). Second countable implies first countable.

For (2), you have identified that every subset is both open and closed. So the space is discrete, and satisfies all separation axioms. Again it is separable because it is countable. A countable basis for the space is just the set of all singletons.