Infinite group with cofinite topology is not a topological group
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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.
|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.|