ONT Re: Topology
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| 1. Topological Spaces
|
| 1.6. Bases and Subbases
|
| In defining the usual topology for the set of real numbers
| we began with the family !B! of open intervals, and from this
| family constructed the topology !T!. The same method is useful
| in other situations and we now examine the construction in detail.
|
| A family !B! of sets is a 'base for a topology' !T!
| iff !B! is a subfamily of !T! and for each point x
| of the space, and each neighborhood U of x, there
| is a member V of !B! such that x in V c U.
|
| Thus the family of open intervals is a base for the usual topology of
| the real numbers, in view of the definition of the usual topology and
| the fact that open intervals are open relative to this topology.
|
| There is a simple characterization of bases which is frequently used
| as a definition: A subfamily !B! of a topology !T! is a base for !T!
| iff each member of !T! is the union of members if !B!. To prove this
| fact, suppose that !B! is a base for the topology !T! and that U in !T!.
| Let V be the union of all members of !B! which are subsets of U and suppose
| that x in U. Then there is W in !B! such that x in W c U, and consequently
| x in V. Hence U c V and since V is surely a subset of U, we have that V = U.
| To show the converse, suppose !B! c !T! and each member of !T! is the union
| of members of !B!. If U in !T!, then U is the union of the members of
| a subfamily of !B!, and for each x in U there is V in !B! such that
| x in V c U. Consequently !B! is a base for !T!.
|
| Although this is a very convenient method for the construction
| of topologies, a little caution is necessary because not every
| family of sets is the base for a topology. For example, let X
| consist of the integers 0, 1, 2, let A consist of 0 and 1, and
| let B consist of 1 and 2. If !S! is the family whose members
| are X, A, B, and the void set, then !S! cannot be the base
| for a topology because: by direct computation, the union
| of members of !S! is always a member, so that if !S! were
| the base of a topology that topology would have to be !S!
| itself, but !S! is not a topology because A |^| B is not
| 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 which is a subset of
| U |^| V. To show the converse, let !B! be a family with the
| specified property and let !T! be the family of all unions
| of members of !B!. A union of members of !T! is itself
| a union of members of !B! and is therefore a member
| of !T!, and it is only necessary to show that the
| intersection of two members U and V of !T! is
| a member of !T!. If x in U |^| V, then we
| may choose U' and V' in !B! such that:
|
| x in W c U' |^| V' c U |^| V.
|
| Consequently U |^| V is the
| union of members of !B!,
| and !T! is a topology.
| þ
|
| JLK, Gen Top, pages 46-47.
|
| John L. Kelley, 'General Topology',
| Van Nostrand Reinhold, New York, NY, 1955.
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