Question:
The number of distinct topologies on the set \(\{1, 2, 3\}\) consisting of exactly four elements is equal to ______. (Answer in integer)
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.
Updated On: Aug 14, 2026
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Verified By Collegedunia
Correct Answer: 9
Solution and Explanation
Step 1: Understand what is being counted.
A topology on \(\{1,2,3\}\) is a collection of subsets of \(\{1,2,3\}\) that contains the empty set \(\varnothing\) and the whole set \(X=\{1,2,3\}\), and is closed under taking unions and intersections of its members. We want topologies with exactly four sets in them. Since \(\varnothing\) and \(X\) are always two of those four sets, we need exactly two more distinct sets \(P\) and \(Q\), each a proper nonempty subset of \(X\), such that \(P \cup Q\) and \(P \cap Q\) also lie inside \(\{\varnothing, X, P, Q\}\).
Step 2: List the possible proper nonempty subsets.
\(X\) has three singletons, \(\{1\}, \{2\}, \{3\}\), and three 2-element sets, \(\{1,2\}, \{1,3\}, \{2,3\}\). We must choose \(P\) and \(Q\) from these six sets.
Step 3: Rule out picking two singletons.
If \(P=\{a\}\) and \(Q=\{b\}\) with \(a \neq b\), then \(P \cup Q = \{a,b\}\), a two-element set not equal to \(\varnothing\), \(X\), \(P\) or \(Q\). Closure under union fails, so no topology of this type exists.
Step 4: Rule out picking two 2-element sets.
If \(P\) and \(Q\) are two different 2-element sets, their intersection is a single element, the one common to both, which is not equal to \(\varnothing\), \(X\), \(P\) or \(Q\). Closure under intersection fails, so no topology of this type exists either.
Step 5: Study the mixed case, one singleton and one 2-element set.
Let \(P=\{a\}\) and let \(Q\) be a 2-element set. Two things can happen:
(i) \(Q\) contains \(a\), so \(Q=\{a,b\}\) for the other element \(b\). Then \(P \cup Q = Q\) and \(P \cap Q = P\), both already in the collection, so closure holds automatically. For each choice of \(a\) (3 ways) there are 2 choices of \(b\), giving \(3 \times 2 = 6\) topologies of this chain type, \(\{\varnothing, \{a\}, \{a,b\}, X\}\).
(ii) \(Q\) does not contain \(a\), so \(Q = X \setminus \{a\}\), the complement of \(P\). Then \(P \cup Q = X\) and \(P \cap Q = \varnothing\), both already in the collection, so closure holds. For each of the 3 choices of \(a\), there is exactly 1 such \(Q\), giving 3 topologies of this partition type, \(\{\varnothing, \{a\}, X\setminus\{a\}, X\}\).
Step 6: Add up all valid topologies.
\[ 6 \text{ (chain type)} + 3 \text{ (partition type)} = 9. \]
Final Answer:
There are 9 distinct topologies on \(\{1,2,3\}\) with exactly four elements. \[ \boxed{9} \]
A topology on \(\{1,2,3\}\) is a collection of subsets of \(\{1,2,3\}\) that contains the empty set \(\varnothing\) and the whole set \(X=\{1,2,3\}\), and is closed under taking unions and intersections of its members. We want topologies with exactly four sets in them. Since \(\varnothing\) and \(X\) are always two of those four sets, we need exactly two more distinct sets \(P\) and \(Q\), each a proper nonempty subset of \(X\), such that \(P \cup Q\) and \(P \cap Q\) also lie inside \(\{\varnothing, X, P, Q\}\).
Step 2: List the possible proper nonempty subsets.
\(X\) has three singletons, \(\{1\}, \{2\}, \{3\}\), and three 2-element sets, \(\{1,2\}, \{1,3\}, \{2,3\}\). We must choose \(P\) and \(Q\) from these six sets.
Step 3: Rule out picking two singletons.
If \(P=\{a\}\) and \(Q=\{b\}\) with \(a \neq b\), then \(P \cup Q = \{a,b\}\), a two-element set not equal to \(\varnothing\), \(X\), \(P\) or \(Q\). Closure under union fails, so no topology of this type exists.
Step 4: Rule out picking two 2-element sets.
If \(P\) and \(Q\) are two different 2-element sets, their intersection is a single element, the one common to both, which is not equal to \(\varnothing\), \(X\), \(P\) or \(Q\). Closure under intersection fails, so no topology of this type exists either.
Step 5: Study the mixed case, one singleton and one 2-element set.
Let \(P=\{a\}\) and let \(Q\) be a 2-element set. Two things can happen:
(i) \(Q\) contains \(a\), so \(Q=\{a,b\}\) for the other element \(b\). Then \(P \cup Q = Q\) and \(P \cap Q = P\), both already in the collection, so closure holds automatically. For each choice of \(a\) (3 ways) there are 2 choices of \(b\), giving \(3 \times 2 = 6\) topologies of this chain type, \(\{\varnothing, \{a\}, \{a,b\}, X\}\).
(ii) \(Q\) does not contain \(a\), so \(Q = X \setminus \{a\}\), the complement of \(P\). Then \(P \cup Q = X\) and \(P \cap Q = \varnothing\), both already in the collection, so closure holds. For each of the 3 choices of \(a\), there is exactly 1 such \(Q\), giving 3 topologies of this partition type, \(\{\varnothing, \{a\}, X\setminus\{a\}, X\}\).
Step 6: Add up all valid topologies.
\[ 6 \text{ (chain type)} + 3 \text{ (partition type)} = 9. \]
Final Answer:
There are 9 distinct topologies on \(\{1,2,3\}\) with exactly four elements. \[ \boxed{9} \]
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