 in !S!.  The reason for this situation is made clear by
| the following theorem.
|
| 11.  Theorem.  A family !B! of sets is a base for some
|      topology for the set X = |_|{B : B in !B!} iff
|      for every two members U and V of !B! and each
|      point x in U |^| V there is W in !B! such
|      that x in W and W c U |^| V.
|
| Proof.  If !B! is a base for some topology, U and V are members
| of !B!, and x is in U |^| V, then, since U |^| V is open, there
| is a member of !B! to which x belongs and w