Hi everyone I'd like to know if the following proof is correct, I think so. And also if there is a more direct approach without the many subcases. Thanks in advance
Let $X$ be an infinite set, and $A_1\subset A_2\subset\ldots\subset A_n\subset \ldots\subset X$. Define the set $A$ as the union of the family $\{A_i\}$. Then $\tau:=\{\varnothing,X,A, A_1,\ldots A_n\ldots\}$ is a topology.
Proof:
Since the empty set and the total space are contained in $\tau$ we only concern to prove that the set is closed under finite intersection and arbitrary unions.
For finite intersection, it is sufficient to prove it for any two elements.It's trivial if one of the elements is either the entire set $X$ or the empty set. So we may assume that neither of them is, only we have two possibilities: If $A_i, A \in \tau$, the intersection certainly is in the set. For $A_i,A_j\in \tau$, so either $i\le j$ or $j\le i$, say the first, then $A_i \subset A_j$ and so $A_i \cap A_j =A_i$, which is in $\tau$.
Finally to prove that the set is closed under arbitrary unions we eliminate the trivial cases (when one member is $X$ or $A$, or even the empty set). We have two cases:
Suppose we have a family of the $A_i$s which is finite, so there must be one $i$ which is the greatest one. Thus the union is the element in the family with the greatest $i$ which is in the set by hypothesis.
On the other hand, if the family is infinite, we claim that the union is $A$. Let $\{A_{j_k}\}$ be an family which is infinite. Clearly the union of these elements is contained in $A$. For the other direction. Let $x$ in A, we have that is in some $n$ such that $x\in A_n$, since $j_n\ge n$. Thus $x\in A_n\subset A_{j_n}\subset \bigcup\{A_{j_k}\}$. Thus $A=\bigcup_k\{A_{j_k}\}$.
$\tau $ is a topology as was to be shown.
EDIT: I´m thinking about and maybe for the last point (arbitrary union of elements is not completely right). This is better
Let $\{B_i\}_{i \in I}$ a family of elements in $\tau$. We may assume that $X$ is not in the family because the result is trivial. Similarly we not consider the cases whenv$A$ and the empty set are in the family. Thus $\{B_i\}_{i \in I} \subset A':=\tau \setminus\{\varnothing, X, A\}$.
Since each family induced a map, then we define $I \overset{f}\rightarrow\tau$; $i\mapsto B_i$. Since $A'$ is countable then $f(I)$ must be either finite or infinite countable. If is finite, one element contained the others and the union must be this element. On the other hand, if is countable, we claim that the union is $A$. To show it, let choose any element in $A$ so there is some $n$ such that $x \in A_n$. But then, there must be some $i$ such that $B_i = A_{n(i)}$ for $n(i)\ge n$ (if not the family cannot be infinite). Thus $$x\in A_n \subset A_{n(i)}=B_i\subset \bigcup \{B_i\}$$
which shows that the union of the family is $A$.
I'd appreciate if some one could see it and verify the proof.
Looks great. If you're looking to prove it from scratch, there's likely not a shorter proof.