Simplicial Structures in Topology

I.1 Topology 15
A
Fig. I.5 Example of a space
that is connected but not
path-connected
space, which implies that B ∪C is connected; moreover, the closure of B ∪C in X
coincides with X and therefore, by Theorem (I.1.18), X is a connected space.
We now prove that X is not path-connected. In order to come to this conclusion,
we shall prove that if f : [0,1] → X isapathsuchthatf (0)=(0, 1 2 ),thenfis the
constant path at the point (0, 1
2 ). Because X ⊂ R2 , we may write f (t)=(x(t),y(t))
for every t ∈ [0,1]; it is not difficult to prove that x and y are continuous functions
(see Exercise 5 on p. 27). We first prove that x is the constant function at 0, that is
to say, x(t)=0foreveryt∈ [0,1]. Indeed, suppose there exists t ′ ∈ [0,1] such that
x(t ′ ) > 0andlett0 = sup{t ∈ [0,1]|x(t)=0}. Since the function x is continuous, we
must have x(t0)=0andt0 < 1 so that we do not contradict the fact that there exists
a t ′ with x(t ′ ) > 0. Since (0, 1 2 ) is the only point with zero for its first coordinate,
we have y(t0)= 1 2 .Sinceyiscontinuous, there exists an ε > 0 with t0 + ε < 1and
such that, for every s ∈ [t0,t0 + ε), y(s) ≥ 1 4 .Sincet0isan upper bound, we can find
t1 ∈ (t0,t0 + ε) such that x(t1) > 0; by Corollary (I.1.16) and because x(t0)=0, we
have x([t0,t1]) ⊇ [0,x(t1)]. Consequently, we are able to find s ∈ (t0,t1) such that
x(s) �= 1
1
n for every integer n > 0(andy(s) ≥ 4 ). Finally, as f (s)=(x(s),y(s)) ∈ X,
it follows that y(s)=0 because the only points of X for which x > 0andx �= 1 n are
of the type (x,0), in contradiction to the fact that y(s) ≥ 1
4 .
We conclude that for every t ∈ [0,1],wehavex(t)=0andsof (t)={(0, 1
2 )};in
other words, f is constant.
(I.1.22) Example. For every n ≥ 1 the unit sphere
S n = {(x0,...,xn) ∈ R n+1 | Σ n i=0x 2 i = 1}
is path connected and consequently, connected. Let a =(x0,...,xn) and b =
(y0,...,yn) be any two points of S n ; we say that b is antipodal to a if yi = −xi
for every i = 0,...,n. Ifb is antipodal to a, the interval R n+1 with end points a and
b goes through the centre (0,...,0). Suppose that b is not antipodal to a; then, for
every t ∈ [0,1],wehave(1 −t)a +tb�= 0. It follows that
f : [0,1] → S n , t ↦→
(1 −t)a + tb
|(1 −t)a + tb|