topology over $X$ Definition. A topology $\tau$ is a collection of open X such that:
(i) $\emptyset$ and $X \in \tau$;
(ii) arbitrary union of $A$ open is opened;
(iii) finite intersection of open is opened.
To prove that the union of topologies is a topology this stems from the topology axiomas, because:
(i) $\emptyset$ and $X\in \cap C$, because $\emptyset$ and $X$ belongs to each topology, so it belongs to the intersection.
(ii) arbitrary intersection of finite intersection is opened. How to prove this fact?
(iii) arbitrary intersection of arbitrary union is opened.How to prove this fact?
Already for the union arbitraria, have:  What's the flaw?
(i) $\emptyset$ and $X\in \cap C$, because $\emptyset$ and $X$ belongs to each topology, so it belongs to the union.
(ii) arbitrary union of finite intersection is opened. How to prove this fact?
(iii) arbitrary union of arbitrary union is opened. How to prove this fact?
Can someone help me please?
 A: The part about $\bigcap C$ being a topology is not too difficult if you write out the definitions and follow your way through.
For an example where $\bigcup C$ is not a topology, take $C = \{\tau_1, \tau_2\}$ where $\tau_i$ comprises all the subsets of $\{1, 2, 3\}$ that are either empty or contain $i$. Then $\{3\} = \{1, 3\} \cap \{2, 3\} \not\in \bigcup C$, but it would have to be if $\bigcup C$ were a topology.
A: Suppose that $\mathscr{C}$ is a family of topologies on $X$. Let $\tau_0=\bigcap\mathscr{C}$. You showed that $\varnothing,X\in\tau_0$. To show that $\tau$ is closed under taking arbitrary unions, let $\mathscr{U}\subseteq\tau$; we want to show that $U\in\tau_0$. Let $\tau\in\mathscr{C}$; then $\mathscr{U}\subseteq\tau_0\subseteq\tau$, so $\mathscr{U}\subseteq\tau$. And $\tau$ is a topology, so $\bigcup\mathscr{U}\in\tau$. Thus, $\bigcup\mathscr{U}\in\tau$ for each $\tau\in\mathscr{C}$, and therefore
$$\bigcup\mathscr{U}\in\bigcap\mathscr{C}=\tau_0\,,$$
as desired.
The proof that $\tau_0$ is closed under taking finite intersections is very similar; see if you can carry it out using this as a model.
To show that $\bigcup\mathscr{C}$ is not necessarily a topology on $X$, you need to produce a counterexample: you want a set $X$ and a family of topologies on $X$ whose union is not a topology on $X$. You can actually do this with just a pair of topologies, say $\tau_1$ and $\tau_2$ on a small set.

*

*Try to find topologies $\tau_1$ and $\tau_2$ on $X=\{0,1,2\}$ such that there are sets $U_1\in\tau_1$ and $U_2\in\tau_2$ whose intersection is not in $\tau_1\cup\tau_2$; this will show that $\tau_1\cup\tau_2$ is not a topology on $X$.

