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