Let $S$ be a set containing exactly one element. Suppose $S = \set x$ for a certain object $x$.  From Topology on Singleton is Indiscrete Topology, the only possible topology on $S$ is the indiscrete topology.  From Union is Idempotent, the union of any number of indiscrete topologies on $S$ is the indiscrete topology. Thus the union of any number of topologies on a set with exactly one element is a topology on that set. {{qed|lemma}} Let $S$ be a set containing exactly two elements. Suppose $S = \set {x, y}$ for certain objects $x$ and $y$.  Then the power set of $S$ is the set: : $\map {\mathcal P} S = \set{\varnothing, \set{x}, \set{y}, \set{x, y}}$ Since all topologies $\tau$ on $S$ are subsets of $\tau \subseteq \mathcal P \left({S}\right)$, one of the following must hold: : $\tau_1 = \O$ : $\tau_2 = \set{\O}$ : $\tau_3 = \set{\set{x}}$ : $\tau_4 = \set{\set{y}}$ : $\tau_5 = \set{\O, \set{x}}$ : $\tau_6 = \set{\O, \set{y}}$ : $\tau_7 = \set{\set{x}, \set{y}}$ : $\tau_8 = \set{\O, \set{x}, \set{y}}$ : $\tau_9 = \set{\set{x}, \set{x, y}}$ : $\tau_{10} = \set{\set{y}, \set{x, y}}$ : $\tau_{11} = \set{\set{x}, \set{y}, \set{x, y}}$ : $\tau_{12} = \set{\set{x, y}}$ : $\tau_{13} = \set{\O, \set{x, y}}$ : $\tau_{14} = \set{\O, \set{x}, \set{x, y}}$ : $\tau_{15} = \set{\O, \set{y}, \set{x, y}}$ : $\tau_{16} = \set{\O, \set{x}, \set{y}, S}$. By definition of a topology, $S$ must be an element of the topology. Thus $\tau_1$ up to $\tau_8$ are not topologies on $S$.  By Empty Set is Element of Topology, for $\tau$ to be a topology for $S$, it is necessary that $\varnothing \in \tau$.  Therefore $\tau_9$ up to $\tau_{12}$ are also not topologies on $S$. By Indiscrete Topology is Topology, $\tau_{13}$ is a topology on $S$. By Discrete Topology is Topology, $\tau_{16}$ is a topology on $S$. By Particular Point Topology is Topology, both $\tau_{14}$ and $\tau_{15}$ are topologies. Finally, it remains to show that any union of elements in $\{\tau_{13}, \tau_{14}, \tau_{15}, \tau_{16}\}$ gives a topology on $S$. By inspection, the following statements are seen to hold: : $\tau_{13} \cup \tau_{14} = \tau_{14}$ : $\tau_{13} \cup \tau_{15} = \tau_{15}$ : $\tau_{13} \cup \tau_{16} = \tau_{16}$ : $\tau_{14} \cup \tau_{15} = \tau_{16}$ : $\tau_{14} \cup \tau_{16} = \tau_{16}$ : $\tau_{15} \cup \tau_{16} = \tau_{16}$ Now let $\paren{\tau_i}_{i \mathop \in I}$ be an arbitrary non-empty indexed set of topologies for a set $S$. By the above, if $\tau_{16}$ is one of the $\tau_i$: : $\displaystyle \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_{16}$ So assume that $\tau_{16}$ is not one of the $\tau_i$. If both $\tau_{14}$ and $\tau_{15}$ are in $\paren{\tau_i}_{i \mathop \in I}$, then by the above: : $\displaystyle \tau = \bigcup_{i \mathop \in I} \tau_i = \tau_{16}$ Now suppose that $\tau_{15}$ is in $\paren{\tau_i}_{i \mathop \in I}$ and $\tau_{14}$ is not. Then by the above it follows that: : $\tau = \tau_{15}$ Now suppose that $\tau_{14}$ is in $\paren{\tau_i}_{i \mathop \in I}$ and $\tau_{15}$ is not. Then by the above it follows that: : $\tau = \tau_{14}$ Finally, assume that $\tau_{14}$ and $\tau_{15}$ are not in $\paren{\tau_i}_{i \mathop \in I}$. Then the only topologies in $\paren{\tau_i}_{i \mathop \in I}$ are $\tau_{13}$.  In the first case, we find: :$\tau = \tau_{13}$ Hence the result. {{qed}}