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.

Applications

• Algebraic group not implies 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.