The number of distinct topologies on the set \(\{1, 2, 3\}\) consisting of exactly four elements is equal to ______. (Answer in integer)
Show Hint
Every topology must contain \(\varnothing\) and \(X\); check which pairs of the remaining proper subsets keep the collection closed under union and intersection.
Step 1: Set up the count as choosing 2 extra sets out of 6 candidates.
Any topology on $\{1,2,3\}$ with exactly four members must contain $\varnothing$ and $X=\{1,2,3\}$, plus two more distinct proper nonempty subsets. There are exactly $6$ proper nonempty subsets of $X$: the three singletons $\{1\},\{2\},\{3\}$ and the three pairs $\{1,2\},\{1,3\},\{2,3\}$. Choosing 2 of these 6 to go with $\varnothing$ and $X$ can be done in $\binom{6}{2}=15$ ways, and each way must be checked for closure under union and intersection.
Step 2: Check the 3 ways of choosing two singletons.
For any two singletons, say $\{1\}$ and $\{2\}$, the union $\{1\}\cup\{2\}=\{1,2\}$ is missing from the collection $\{\varnothing,\{1\},\{2\},X\}$. All 3 such choices fail the same way, so none of them give a valid topology.
Step 3: Check the 3 ways of choosing two pairs.
For any two of the 2-element sets, say $\{1,2\}$ and $\{1,3\}$, the intersection $\{1,2\}\cap\{1,3\}=\{1\}$ is missing from the collection $\{\varnothing,\{1,2\},\{1,3\},X\}$. All 3 such choices fail the same way, so none of them work either.
Step 4: Check the remaining 9 mixed choices, one singleton and one pair.
There are $3\times 3=9$ ways to pick one singleton and one pair. Splitting these by whether the singleton sits inside the pair: if it does, for example $\{1\}$ with $\{1,2\}$, the union is $\{1,2\}$ (already present) and the intersection is $\{1\}$ (already present), so this is valid; there are 6 such combinations. If the singleton is disjoint from the pair, for example $\{1\}$ with $\{2,3\}$, the union is $X$ (already present) and the intersection is $\varnothing$ (already present), so this is valid too; there are 3 such combinations. All 9 mixed choices pass the closure check.
Step 5: Total the valid combinations.
Out of the $15$ possible choices, $0$ come from two singletons, $0$ from two pairs, and $9$ from the mixed case, giving a total of $9$ topologies with exactly four elements.
\[ \boxed{9} \]
