Simplicial Structures in Topology

4 I Fundamental Concepts
x
z
y
Fig. I.1
The topology U generated by a basis B may be described in another way:
(I.1.2) Lemma. Let X be a set, let B be a basis of open sets of X, and let U be the
topology generated by B. Then, U coincides with the set of all unions of elements
of B.
Proof. Given any set {Uα | α ∈ J, Uα ∈ B} of open sets of the basis B, since all
elements B in are open sets and U is a topology, the union �
α∈J Uα is also open.
Conversely, given U ∈ U, for each x ∈ U, there exists Bx ∈ B such that x ∈ Bx ⊂ U.
Therefore, U = �
x∈U Bx,thatistosay,Uisa union of elements of B. �
We now look into the concept of sub-basis. A set of subsets S of X is a subbasis
for X if the union of all elements of S coincides with X (in other words, if
property B1, given above, holds). The next result provides a good reason to work
with sub-bases.
(I.1.3) Theorem. Let S be a sub-basis of a set X. Then, the set U of all unions of
finite intersections of elements of S is a topology.
Proof. It is enough to prove that the set B of all finite intersections of elements of
S is a basis and then apply Lemma (I.1.2).
B1: For each x ∈ X there exists B ∈ S such that x ∈ B;however,B∈B, since all
elements of S are elements of B.
B2: Given B1 = �n i=1C1 i and B2 = �m j=1C 2 j in B, the property holds because
� � � �
n�
m�
B1 ∩ B2 =
C
i=1
1 i
∩
C
j=1
2 j
∈ B. �
Clearly, any basis is a sub-basis. For instance, the set {{x}|x ∈ X} is a subbasis
for the discrete topology on X as well as a basis for that same topology. On
the contrary, the family of all open intervals of R with length 1 is an example of a
sub-basis that is not a basis. Property B1 holds but not property B2. The basis it
generates consists of all open intervals of R and we have, therefore, the Euclidean
topology.
Let X and Y be two topological spaces with respective topologies U and V. The
product topology of the set X ×Y is generated by the basis
B = {U ×V |U ∈ U , V ∈ U}.