Simplicial Structures in Topology

14 I Fundamental Concepts
We now define the function f : [0,1] → X
�
f0(2t) if 0 ≤ t ≤
(∀t ∈ [0,1]) f (t)=
1 2
f1(2 − 2t) if 1
2 ≤ t ≤ 1.
Since f ( 1
2 )= f0(1)= f1(0)=a, it follows from Corollary (I.1.10) that the function
f is continuous. However, f (0)=x0, f (1)=x1 and so f is a path that links x0 to x1.
3. Given (x1,y1),(x2,y2) ∈ X ×Y, choose two paths
fX : [0,1] → X such that fX(0)=x1 , fX (1)=x2
fY : [0,1] → Y such that fY (0)=y1 , fY (1)=y2.
The path ( fX, fY ): [0,1] → X ×Y t↦→ ( fX (t), fY (t)) links (x1,y1) to (x2,y2) and
X ×Y is, therefore, path-connected. �
The Euclidean space R n is path-connected for every n > 0. Indeed, given
x0,x1 ∈ R n ,wedefine
f : [0,1] → R n , (∀t ∈ [0,1]) f (t)=tx1 +(1 − t)x0.
In particular, every interval of R is path-connected (as well as any convex subspace
of R n ).
(I.1.21) Theorem. Any path-connected space is connected.
Proof. Suppose X to be the union of two non-empty, disjoint subspaces U, V which
are simultaneously open and closed. Take two points x0,x1 ∈ X where x0 ∈ U and
x1 ∈ V. SinceX is path-connected, there exists a map
Then,
f : [0,1] → X , f (0)=x0 , f (1)=x1.
(U ∩ f ([0,1])) ∩ (V ∩ f ([0,1])) �= /0 ,
contradicting the fact that f ([0,1]) is connected (see Theorem (I.1.13)). �
In general, it is not true that a connected space is path-connected; here is an
example.
Consider the following sets of points from the Euclidean plane:
��
A = 0, 1
��
,
2
��
1
B =
n ,t
�
�
| n ∈ N and t ∈ [0,1] ,
C = {(t,0) |t ∈ (0,1]} =(0,1] ×{0}
and endow X = A ∪ B ∪C with the topology induced by the Euclidean topology of
R 2 , as shown in Fig. I.5. It is immediate to verify that B ∪C is a path-connected