Suppose a setis given. To promote it to a topological spacewe must describe a family of subsets ofwhich satisfy certain conditions. Of the many ways which define such a family of subsets, one is using a functionwhich sends every subsetofto its closure

The closureofmust satisfy these conditions

With the closure defined as above, we can write down the theorem:

Letrepresent a set and letrepresent a function such that

1.

2. for each

3.for

4.for each

Then the familyis a topology onwhereis the discrete topology onandfor each

Proof

Ifthen(1)

In fact ifthenand

Ifthen sincewe havehenceAlso since

Ifthenand any finite intersection is in

Letthenfor eachBy (1) we obtainfor each

Henceandwith

The union of any subset ofmust be inBy usingas the topology onwe show thatIn factbecause eachis closed andsoSimilarly

Sinceis open infor some

Sincewe obtain