Simplicial Structures in Topology

2 I Fundamental Concepts
C1 /0,X ∈ C;
C2 if {Uα | α = 1,...,n} is any finite set of elements of C,then
n�
Uα ∈ C
α=1
(in other words, any finite union of closed sets is a closed set);
C3 if {Uα | α ∈ J} is any set of elements of C,then
�
Uα ∈ C
α∈J
(in other words, any intersection of closed sets is a closed set).
For any space X,theclosure Y of a subset Y ⊂ X is the intersection of all closed
subsets of X which contain Y; it is, obviously, the smallest closed subset containing
Y, andY∈C. Theinterior ˚Y of Y is the union of all open subsets of X contained
in Y ; it is the largest open set contained in Y ,and˚Y ∈ U. The following lemma is a
useful characterization of the closure of a set.
(I.1.1) Lemma. Let Y ⊂ X; theny∈ Y if and only if every open set U containing y
intersects Y .
Proof. In fact, if U ∩Y = /0, then X �U is a closed subset of X which contains Y ;
since y belongs to the closure of Y, it follows that y ∈ X �U, a contradiction.
Conversely, suppose that there exists a closed set C ⊂ X containing Y and which
does not contain y. ThenX �C is an open set, y ∈ X �C, and(X �C) ∩Y = /0, a
contradiction. �
For any set X, there are two topologies that come immediately to mind:
1. The discrete topology whose open sets are all the subsets of X, thatistosay,
Ud = P(X)=2X .
2. The trivial topology with Ub = {/0,X}.
Clearly, Ud ⊃ Ub and this fact brings to mind the idea of comparing two topologies
on the same set: we shall say that topology U ′ is finer than topology U if U ′ ⊃ U;
hence, the discrete topology of X is finer than the trivial one.
AsetBof subsets of X is a basis for a topology on X or a basis of open sets for
X if:
B1 For each x ∈ X,thereisaB∈Bsuch that x ∈ B (that is to say, X = �
B∈B B).
B2 If B1,B2 ∈ B and B1 ∩B2 �= /0, then for each x ∈ B1 ∩B2 there exists B3 ∈ B
such that x ∈ B3 ⊂ B1 ∩ B2.
AbasisBgenerates a topology U on X automatically, by requiring that U ⊂ X
be open in the topology U if, and only if, for each x ∈ U, there exists B ∈ U such that
x ∈ B ⊂ U; formally,
U ∈ U ⇐⇒ (∀x ∈ U)(∃B ∈ B) x ∈ B ⊂ U.