Simplicial Structures in Topology

I.1 Topology 17
with the topology induced by the Euclidean topology of R, is compact; on the other
hand, the space R is not compact: indeed, the open covering
of R has no finite subcovering.
U = {(n,n + 2) | n ∈ Z}
(I.1.23) Theorem. Any closed subspace of a compact space is compact.
Proof. Let Y be a closed subspace of a compact space X. Let C = {U j | j ∈ J}
be a covering of Y such that, for every j ∈ J, Uj is an open set of X; then the set
C∪{X �Y} is an open covering of X. SinceX is compact, it has a finite subcovering
U1 ∪ ...∪Un ∪ (X �Y)=X.
As no x ∈ Y is in X �Y, {U1,...,Un} is a covering of Y ,andso,Y is compact. �
We now introduce a special type of topological space which is needed for the
next result. We say that a topological space X is Hausdorff, if for every two distinct
points x,y of X, we can find two disjoint open sets Ux and Uy of X such that x ∈ Ux,
y ∈ Uy.
It is easily proved that a subspace of a Hausdorff space is also Hausdorff; and
it is easily seen that the Euclidean space Rn is Hausdorff. Here is an example of a
space which is not Hausdorff.
(I.1.24) Example. Let Y = R ∪{∗} be the set given by the union of the real line R
and an external point ∗; for the set of open subsets of Y , we choose the set consisting
of the empty set and all subsets of Y which are the union of an open set of R and
the point ∗ (it is left to the reader to verify that these sets define a topology on Y).
The space Y here defined is not Hausdorff because the intersection of any two nonempty,
open sets of Y is not empty!
It follows from the definition that in any Hausdorff space every point is closed.
More generally, all compact subspaces of a Hausdorff space are closed, as stated in
the next theorem.
(I.1.25) Theorem. Any compact subspace of a Hausdorff space is closed.
Proof. Let Y be a compact subspace of a Hausdorff space X. Wemustprovethat
X �Y is open. Let us take any x ∈ X �Y; sinceX is Hausdorff, for every y ∈ Y, let
us take two open sets Uy,Vy of X such that
y ∈ Uy , x ∈ Vy , Uy ∩Vy = /0.
The set U = {Uy | y ∈ Y } is a covering of Y by open sets of X. SinceY is compact,
Y is covered by a finite number of these open sets, for instance, Y ⊂ � n i=1 Ui. Letus
take the set of open sets {V1,...,Vn} corresponding to this finite covering of Y and
consider the open set V = � n j=1 Vj. It is not difficult to prove that V is an open set
such that x ∈ V ⊂ X �Y. �