Infinite group with cofinite topology is not a topological group

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Statement

Let G be an infinite group. Equip G with the cofinite topology. With this topology, G is not a topological group.

Applications

Proof

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Given: An infinite group G equipped with the cofinite topology.

To prove: The multiplication map G \times G \to G is not continuous where G \times G is equipped with the product topology.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 A basis for the product topology on G \times G is given by subsets of the form U \times V where U,V are cofinite subsets of G. Definition of product topology G has the cofinite topology direct (note that we can exclude the products involving empty open subsets from the basis).
2 Any two cofinite subsets of G have a non-empty intersection G is infinite The intersection of two cofinite subsets has complement equal to the union of their complements, which is a union of two finite subsets, hence finite. Since G is infinite, it cannot be everything, so the intersection is non-empty.
3 For any (possibly equal, possibly distinct) cofinite subsets U,V of G, U \cap V^{-1} is non-empty. Step (2) G is a group, so the inverse map, being a bijection, preserves cofiniteness. In particular, V^{-1} is also cofinite, and Step (2) applies.
4 For any (possibly equal, possibly distinct) cofinite subsets U,V of G, there exists g \in G such that (g,g^{-1}) \in U \times V. Step (3) Pick g \in U \cap V^{-1} -- the intersection is non-empty by Step (3).
5 Consider the multiplication map G \times G \to G. Under this map, the inverse image of the identity element of G is the set \{ (g,g^{-1}) \mid g \in G \}. direct from definition of group
6 The inverse image of the identity element under the multiplication map G \times G \to G intersects every basis open subset of G \times G. Steps (1), (4), (5) Step-combination direct
7 The inverse image of the identity element under the multiplication map G \times G \to G is not a closed subset of G \times G. Step (6) For a subset to be closed, its complement must be a union of basis open subsets. But Step (6) rules out that possibility (note that the edge case of the complement being empty need not concern us here).
8 The multiplication map G \times G \to G is not continuous where G \times G is given the product topology. G has the cofinite topology, so in particular singleton subsets are closed. Step (7) The singleton subset comprising the identity element is closed in G. If the multiplication map were continuous, its inverse image would be closed in G \times G. But Step (7) says the inverse image is not closed, so the multiplication map is not continuous.