Theorem
Given a topological space (X,T) with an equivalence relation R, the quotient set X/R is a topology.
Proof
Define the set of all open sets of![]()
Define a mapping![]()
Then
and![]()
is a topological space so
hence![]()
If
then
and
are open sets in
hence
but
Hence
is open in![]()
Let
be a family of open sets in
then
and
Hence
is an open subset of![]()
The collection of open sets of
forms a topology on![]()