Hi everyone I'd like to know if the following is correct. I'd appreciate any suggestion. Thanks in advance.
From Dudley´s book: Let $A_n$ be the set of all the integers greater than $n$. Let $B_n=2^{\{1,\ldots n\}}$. Let $\mathcal{T}_n$ be the collection of sets of positive integers that are either in $B_n$ or of the form $A_n\cup B$ for some $B\in B_n$. Prove that: $\mathcal{T}_n$ is topology, $\mathcal{T}_n$ for $n=1,2\ldots $ is an inclusion-chain of topologies whose union is not a topology. Describe the smallest topology which includes $\mathcal{T}_n$ for all $n$.
Sketch proof: We consider $\mathbb{N}$ as the positive integers. Let $n\in \mathbb{N}$, then is clear that $\varnothing \in \mathcal{T}_n$ as the empty union, and also $\mathbb{N}=A_n\cup \{1,\ldots n\}\in \mathcal{T}_n$.
Let $U,V\in \mathcal{T}_n$, if $U$ and $V$ are in $B_n$, clearly $U\cap V \in B_n$. If $V=A_n\cup B$, for $B\in B_n$, and $U$ is as above, $V\cap U=(A_n\cup B')\cap U=U\cap B´$. Now if $U=A_n\cup B$ and $V=A_n\cup B´$ for $B,B´\in B_n$, we have $U\cap V=A_n \cup B\cap B´$. Thus, in any case the intersection is in $\mathcal{T}_n$.
Let $\mathcal{U}\subset \mathcal {T}_n$. Then, if all the elements in $\mathcal{U}$ are in $B_n$, the union as well. If there is elements of the form $A_n\cup B$. We define $\mathcal{U}^*=\{A_n\cup {u}: u\in \mathcal{U}\}$. Thus, $$\bigcup \mathcal{U}=\bigcup \mathcal{U}^*=A_n\cup B^*$$ where $B^*$ is the union of the $B´s$ which are in $B_n$. $\square$
Let $n<m\in \mathbb{N}$. Then, if $B\in B_n$ certainly $B$ is in $B_m$ and so $B\in \mathcal{T}_m$. On the other hand, if $A_n\cup B=A_m\cup (A_n\setminus A_m\cup B)$ and $(A_n\setminus A_m\cup B)\in B_m$. Hence $\mathcal{T}_n\subset \mathcal{T}_m$. Now we define $\mathscr{T}=\bigcup_i \mathcal{T}_i$. So $\{2k\}\in\mathscr{T}$ for all $k$, but $$\bigcup\{2k:k\in \mathbb{N}\}=2\mathbb{N}\notin \mathscr{T}$$ Let $\mathcal{G}=\bigcap\{\mathscr{T}\subset \mathscr{S}: \mathscr{S} \text{ is a topology in }\mathbb{N}\}$. Since for all singletons, $\{n\} \in \mathscr{T}$, it follows that any topology which contains $\mathscr{T}$ must contain $\{n\}$, i.e., $\{n\}\in \mathcal{G}$, which proves that $\mathcal{G}= 2^\mathbb{N}$. $\square$
Jose, I read through everything, and I'm not seeing any mistakes. You might want to mention that $i$ is a natural number when you write $\mathcal{T}= \bigcup_i \mathcal{T}_i$
Cheers,
-Russ