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neighborhood system on a set∗
CWoo†
2013-03-21 22:02:01
In point-set topology, a neighborhood system is defined as the set of neighborhoods of some point in the topological space.
However, one can start out with the definition of a “abstract neighborhood
system” N on an arbitrary set X and define a topology T on X based on this
system N so that N is the neighborhood system of T . This is done as follows:
Let X be a set and N be a subset of X × P (X), where P (X) is the
power set of X. Then N is said to be a abstract neighborhood system
of X if the following conditions are satisfied:
1. if (x, U ) ∈ N, then x ∈ U ,
2. for every x ∈ X, there is a U ⊆ X such that (x, U ) ∈ N,
3. if (x, U ) ∈ N and U ⊆ V ⊆ X, then (x, V ) ∈ N,
4. if (x, U ), (x, V ) ∈ N, then (x, U ∩ V ) ∈ N,
5. if (x, U ) ∈ N, then there is a V ⊆ X such that
• (x, V ) ∈ N, and
• (y, U ) ∈ N for all y ∈ V .
In addition, given this N, define the abstract neighborhood system
around x ∈ X to be the subset Nx of N consisting of all those
elements whose first coordinate is x. Evidently, N is the disjoint
union of Nx for all x ∈ X. Finally, let
T
=
{U ⊆ X | for every x ∈ U , (x, U ) ∈ N}
=
{U ⊆ X | for every x ∈ U , there is a V ⊆ U , such that (x, V ) ∈ N}.
The two definitions are the same by condition 3. We assert that T defined
above is a topology on X. Furthermore, Tx := {U | (x, U ) ∈ Nx } is the set of
neighborhoods of x under T .
∗ hNeighborhoodSystemOnASeti created: h2013-03-21i by: hCWooi version: h38905i
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Proof. We first show that T is a topology. For every x ∈ X, some U ⊆ X, we
have (x, U ) ∈ N by condition 2. Hence (x, X) ∈ N by condition 3. So X ∈ T .
Also, ∅ ∈ T is vacuously satisfied, for no x ∈ ∅. If U, V ∈ T , then U ∩ V ∈ T by
condition S
4. Let {Ui } be a subset of T whose elements are indexed by I (i ∈ I).
Let U = Ui . Pick any x ∈ U , then x ∈ Ui for some i ∈ I. Since Ui ∈ T ,
(x, Ui ) ∈ N. Since Ui ⊆ U , (x, U ) ∈ N by condition 3, so U ∈ T .
Next, suppose N is the set of neighborhoods of x under T . We need to show
N = Tx :
1. (N ⊆ Tx ). If N ∈ N , then there is U ∈ T with x ∈ U ⊆ N . But
(x, U ) ∈ N, so by condition 3, (x, N ) ∈ N, or (x, N ) ∈ Nx , or N ∈ Tx .
2. (Tx ⊆ N ). Pick any U ∈ Tx and set W = {z | U ∈ Tz }. Then x ∈ W ⊆ U
by condition 1. We show W is open. This means we need to find, for each
z ∈ W , a V ⊆ W such that (z, V ) ∈ N. If z ∈ W , then (z, U ) ∈ N. By
condition 5, there is V ∈ N such that (z, V ) ∈ N, and for any y ∈ V ,
(y, U ) ∈ N, or y ∈ U by condition 1. So y ∈ W by the definition of W , or
V ⊆ W . Thus W is open and U ∈ N .
This completes the proof. By the way, W defined above is none other than the
interior of U : W = U ◦ .
Remark. Conversely, if T is a topology on X, we can define Nx to be the
set consisting of (x, U ) such that U is a neighborhood of x. The the union N of
Nx for each x ∈ X satisfies conditions 1 through 5 above:
1. (condition 1): clear
2. (condition 2): because (x, X) ∈ N for each x ∈ X
3. (condition 3): if U is a neighborhood of x and V a supserset of U , then V
is also a neighborhood of x
4. (condition 4): if U and V are neighborhoods of x, there are open A, B
with x ∈ A ⊆ U and x ∈ B ⊆ V , so x ∈ A ∩ B ⊆ U ∩ V , which means
U ∩ V is a neighborhood of x
5. (condition 5): if U is a neighborhood of x, there is open A with x ∈ A ⊆ U ;
clearly A is a neighborhood of x and any y ∈ A has U as neighborhood.
So the definition of a neighborhood system on an arbitrary set gives an alternative way of defining a topology on the set. There is a one-to-one correspondence between the set of topologies on a set and the set of abstract neighborhood
systems on the set.
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