# Infinite group with cofinite topology is not a topological group

From Groupprops

## Statement

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

## Applications

## Proof

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

**To prove**: The multiplication map is not continuous where 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 is given by subsets of the form where are cofinite subsets of . | Definition of product topology | 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 have a non-empty intersection | 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 is infinite, it cannot be everything, so the intersection is non-empty. | ||

3 | For any (possibly equal, possibly distinct) cofinite subsets of , is non-empty. | Step (2) | is a group, so the inverse map, being a bijection, preserves cofiniteness. In particular, is also cofinite, and Step (2) applies. | ||

4 | For any (possibly equal, possibly distinct) cofinite subsets of , there exists such that . | Step (3) | Pick -- the intersection is non-empty by Step (3). | ||

5 | Consider the multiplication map . Under this map, the inverse image of the identity element of is the set . | direct from definition of group | |||

6 | The inverse image of the identity element under the multiplication map intersects every basis open subset of . | Steps (1), (4), (5) | Step-combination direct | ||

7 | The inverse image of the identity element under the multiplication map is not a closed subset of . | 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 is not continuous where is given the product topology. | has the cofinite topology, so in particular singleton subsets are closed. | Step (7) | The singleton subset comprising the identity element is closed in . If the multiplication map were continuous, its inverse image would be closed in . But Step (7) says the inverse image is not closed, so the multiplication map is not continuous. |