Prob. 4(a), Sec. 13 in Munkres' TOPOLOGY, 2nd ed: Any necessary and sufficient conditions for a union of topologies to be a topology? Let $X$ be a non-empty set, let $$\left\{ \ \mathscr{T}_\alpha \ \colon \ \alpha \in J \ \right\}$$ be a non-empty collection of topologies on $X$, and let $$ \mathscr{T} \colon= \bigcup_{\alpha\in J} \mathscr{T}_\alpha.$$
Then $\mathscr{T}$ may or may not be a topology on $X$. 
What is (are) the necessary and sufficient condition(s) for $\mathscr{T}$ to be a topology on $X$? 
As a sufficient condition, I can think of the following: 
For any $\alpha$ and $\beta \in J$ such that $\alpha \neq \beta$, we must have either $\mathscr{T}_\alpha \subset \mathscr{T}_\beta$ or  $\mathscr{T}_\beta \subset \mathscr{T}_\alpha$. 
Am I right? 
Is the above condition necessary also? 
Can we formulate other conditions that are either necessary or sufficient, or both for $\mathscr{T}$ to be a topology? 
Can we come up with any condition(s) on the set $X$ that turn out to be either necessary or sufficient, or both, for $\mathscr{T}$ to be a topology? 
 A: Since you asked me to answer here: http://dbfin.com/topology/munkres/chapter-2/section-13-basis-for-a-topology/problem-5-solution/#comment-2456653263.
Your sufficient condition is not in fact sufficient
Indeed, let $X=\mathbb{R}$, and consider topology $\mathcal{T}_n=\{\emptyset,X,I_1,\ldots,I_n=(-1,1-\tfrac{1}{n})\}$. Each $\mathcal{T}_n$ is clearly a topology on $\mathbb{R}$, but their union contains every $I_n$, but not their countable union $\cup_n I_n = (-1,1)$.
A general (not very useful) necessary and sufficient condition
So, given an indexed family of topologies $\{\mathcal{T}_\alpha\}_{\alpha\in J}$ on $X$, the question is, when $\mathcal{T}=\cup_{\alpha\in J} \mathcal{T}_\alpha$ is a topology. Let us check all conditions.


*

*The empty set and $X$ are in $\mathcal{T}$ because they are in each topology.

*Given $U_\beta\in\mathcal{T}$, $\beta\in K$, we need $\cup_{\beta\in K} U_\beta\in \mathcal{T}$.
For every $\alpha\in J$, let $S(\alpha)$ be the set of indexes $\beta$ such that $U_\beta\in \mathcal{T}_\alpha$. Then, $\cup_{\beta\in K}U_\beta = \cup_{\alpha\in J}\cup_{\beta\in S(\alpha)}U_\beta$, but $\cup_{\beta\in S(\alpha)}U_\beta\in \mathcal{T}_\alpha$.


*Similarly for (finite) intersections, $\cap_{n=1}^N U_n = \cap_{\alpha\in J}\cap_{\beta\in S(\alpha)}U_\beta$, but $\cap_{\beta\in S(\alpha)}U_\beta\in \mathcal{T}_\alpha$ because $\cup_\alpha S(\alpha)=\{1,\ldots,N\}$ is finite.


Therefore, we conclude, that a sufficient condition is as follows:
arbitrary unions and finite intersections of sets
from pairwise different topologies must belong to
some topologies in the family.

This condition is also, clearly, necessary.
This condition is not very useful, but about to be the best one can find in general. It can also be useful in specific cases, such as those below.
A specific trivial case
If $X$ has two elements, then the union of an arbitrary family of topologies on $X$ is a topology on $X$.
Finite case
Suppose now $X$ is arbitrary, but the family of topologies contains a finite number of topologies, $\mathcal{T}_n$, $1\le n\le N$. Then, we have the following criterion: $\mathcal{T}=\cup_n\mathcal{T}_n$ is a topology iff for every $U\in\mathcal{T}_i$ and $V\in\mathcal{T}_j$, $U\cup V$ and $U\cap V$ belong to some $\mathcal{T}_k$ and $\mathcal{T}_{k'}$, respectively.
Note, that this condition holds for the above example of topologies on $\mathbb{R}$, even though their union is not a topology. This is, of course, because this criterion is for a finite number of topologies only, and there we have a countable number of topologies.
A: Simply put, a "necessary and sufficient condition for" some statement $\;P\;$ is a statement that is equivalent to $\;P\;$.  And presumably that equivalent statement is simpler, or more useful, or more explicit, than $\;P\;$.
For this specific question, let's try the calculational approach.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
\newcommand{\a}{\alpha}
\newcommand{\b}{\beta}
\newcommand{\c}{\gamma}
\newcommand{\V}{\mathcal V}
\newcommand{\W}{\mathcal W}
\newcommand{\T}[1]{{\mathcal T}_{#1}}
\newcommand{\istop}[2]{#1\text{ is a topology on }#2}
$
We are given a non-empty set $\;X\;$, and a non-empty set $\;J\;$ which is used as an 'index set'.  To avoid visual noise, we will implicitly assume that $\;\a,\b,\c \in J\;$.  We are given a family $\;\T{-}\;$ where for every $\;\alpha\;$, $\;\istop{\T{\a}}{X}\;$.
The question is now: find an equivalent for the statement
$$
\tag{0}
\istop{\langle \cup \a :: \T{\a} \rangle}{X}
$$
Let's also make explicit the definition of a topology:
\begin{align}
\tag{1}
& \istop{\V}{X} \;\equiv\;
\\ & \qquad
\tag{1a}
X \in \V
\\ & \qquad \land\;
\tag{1b}
\langle \forall A,B : A \in \V \land B \in \V : A \cap B \in \V \rangle
\\ & \qquad \land\;
\tag{1c}
\langle \forall \W : W \subseteq \V : \bigcup \W \in \V \rangle
\end{align}
Now, let's try to calculate an equivalent for $\ref{0}$.

Using definition $\ref{1}$ we can write $\ref{0}$ as $\ref{0a} \land \ref{0b} \land \ref{0c}$, and we will rewrite those three parts in turn.
First part $\ref{a}$ of the definition:
$$\calc
    \tag{0a}
    X \in \langle \cup \a :: \T{\a} \rangle
\op=\hint{definition of $\;\cup\;$}
    \langle \exists \a :: X \in \T{\a} \rangle
\op=\hint{each $\;\T{\a}\;$ is a topology, using $\ref{1a}$}
    \langle \exists \a :: \true \rangle
\op=\hint{the index set $\;J\;$ is not empty}
    \true
    \tag{A}
\endcalc$$
So this part is always satisfied.
For part $\ref{b}$ we get:
$$\calc
    \tag{0b}
    \langle \forall A,B : A \in \langle \cup \a :: \T{\a} \rangle \land B \in \langle \cup \a :: \T{\a} \rangle : A \cap B \in \langle \cup \a :: \T{\a} \rangle \rangle
\op=\hint{definition of $\;\cup\;$, three times; rename dummies}
    \langle \forall A,B : \langle \exists \a :: A \in \T{\a} \rangle \land \langle \exists \b :: B \in \T{\b} \rangle : \langle \exists \c :: A \cap B \in \T{\c} \rangle \rangle
\op=\hint{logic: $\;\langle \forall x : \langle \exists y :: P \rangle : Q \rangle\;$ is equivalent to $\;\langle \forall x,y : P : Q \rangle\;$, twice}
    \langle \forall A,B,\a,\b : A \in \T{\a} \land B \in \T{\b} : \langle \exists \c :: A \cap B \in \T{\c} \rangle \rangle
    \tag{B}
\endcalc$$
In words: For any two open sets in any of the topologies (from the family), their intersection is in some topology (in the family).
Finally part $\ref{c}$, where we can do even less:
$$\calc
    \tag{0c}
    \langle \forall \W : \W \subseteq \langle \cup \a :: \T{\a} \rangle : \bigcup \W \in \langle \cup \a :: \T{\a} \rangle \rangle
\op=\hint{definition of $\;\subseteq$; definition of $\;\cup\;$, twice}
    \langle \forall \W : \langle \forall Z : Z \in \W : \langle \exists \a :: Z \in \T{\a} \rangle \rangle : \langle \exists \a :: \bigcup \W \in \T{\a} \rangle \rangle
    \tag{C}
\endcalc$$
In words: For any collection of sets, where each is open in some topology (in the family), the union of those sets is in some topology (in the family).
Combining the above, we've proven that $\ref{B} \land \ref{C}$ is an equivalent, and more explicit, version of $\ref{0}$.

Without more knowledge about the family of topologies $\;\T{-}\;$, it is not possible to go any further.
The only thing one could still do is to rewrite the last line of part $\ref{c}$ above, as follows:
$$\calc
    \tag{C}
    \langle \forall \W : \langle \forall Z : Z \in \W : \langle \exists \a :: Z \in \T{\a} \rangle \rangle : \langle \exists \a :: \bigcup \W \in \T{\a} \rangle \rangle
 \op=\hints{set theory: introduce a choice function}
       \hint{-- to convert the nested $\;\langle \forall \ldots \rangle\;$ to a $\;\langle \exists \ldots \rangle\;$}
     \langle \forall \W : \langle \exists f : f \in \W \to J : \langle \forall Z : Z \in \W : Z \in \T{f(Z)} \rangle \rangle : \langle \exists \a :: \bigcup \W \in \T{\a} \rangle \rangle 
\op=\hint{logic: $\;\langle \forall x : \langle \exists y :: P \rangle : Q \rangle\;$ is equivalent to $\;\langle \forall x,y : P : Q \rangle\;$}
     \langle \forall \W, f : f \in \W \to J \;\land\; \langle \forall Z : Z \in \W : Z \in \T{f(Z)} \rangle : \langle \exists \a :: \bigcup \W \in \T{\a} \rangle \rangle
\endcalc$$
But that does not seem useful in most contexts...
