# Infinite group with cofinite topology is not a topological group

## Statement

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

## 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.