[<< wikibooks] Topology/Path Connectedness
== Definition ==
A topological space

X

{\displaystyle X}
is said to be path connected if for any two points

x

0

,

x

1

∈
X

{\displaystyle x_{0},x_{1}\in X}
there exists a continuous function

f
:
[
0
,
1
]
→
X

{\displaystyle f:[0,1]\to X}
such that

f
(
0
)
=

x

0

{\displaystyle f(0)=x_{0}}
and

f
(
1
)
=

x

1

{\displaystyle f(1)=x_{1}}

== Example ==
All convex sets in a vector space are connected because one could just use the segment connecting them, which is

f
(
t
)
=
t

a
→

+
(
1
−
t
)

b
→

{\displaystyle f(t)=t{\vec {a}}+(1-t){\vec {b}}}
.
The unit square defined by the vertices

[
0
,
0
]
,
[
1
,
0
]
,
[
0
,
1
]
,
[
1
,
1
]

{\displaystyle [0,0],[1,0],[0,1],[1,1]}
is path connected. Given two points

(

a

0

,

b

0

)
,
(

a

1

,

b

1

)
∈
[
0
,
1
]
×
[
0
,
1
]

{\displaystyle (a_{0},b_{0}),(a_{1},b_{1})\in [0,1]\times [0,1]}
the points are connected by the function

f
(
t
)
=
[
(
1
−
t
)

a

0

+
t

a

1

,
(
1
−
t
)

b

0

+
t

b

1

]

{\displaystyle f(t)=[(1-t)a_{0}+ta_{1},(1-t)b_{0}+tb_{1}]}
for

t
∈
[
0
,
1
]

{\displaystyle t\in [0,1]}
.The preceding example works in any convex space (it is in fact almost the definition of a convex space).

Let

X

{\displaystyle X}
be a topological space and let

a
,
b
,
c
∈
X

{\displaystyle a,b,c\in X}
. Consider two continuous functions

f

1

,

f

2

:
[
0
,
1
]
→
X

{\displaystyle f_{1},f_{2}:[0,1]\to X}
such that

f

1

(
0
)
=
a

{\displaystyle f_{1}(0)=a}
,

f

1

(
1
)
=
b
=

f

2

(
0
)

{\displaystyle f_{1}(1)=b=f_{2}(0)}
and

f

2

(
1
)
=
c

{\displaystyle f_{2}(1)=c}
. Then the function defined by

f
(
x
)
=

{

f

1

(
2
x
)

if

x
∈
[
0
,

1
2

]

f

2

(
2
x
−
1
)

if

x
∈
[

1
2

,
1
]

{\displaystyle f(x)=\left\{{\begin{array}{ll}f_{1}(2x)&{\text{if }}x\in [0,{\frac {1}{2}}]\\f_{2}(2x-1)&{\text{if }}x\in [{\frac {1}{2}},1]\\\end{array}}\right.}

Is a continuous path from

a

{\displaystyle a}
to

c

{\displaystyle c}
. Thus, a path from

a

{\displaystyle a}
to

b

{\displaystyle b}
and a path from

b

{\displaystyle b}
to

c

{\displaystyle c}
can be adjoined together to form a path from

a

{\displaystyle a}
to

c

{\displaystyle c}
.

== Relation to Connectedness ==
Each path connected space

X

{\displaystyle X}
is also connected. This can be seen as follows:
Assume that

X

{\displaystyle X}
is not connected. Then

X

{\displaystyle X}
is the disjoint union of two open sets

A

{\displaystyle A}
and

B

{\displaystyle B}
. Let

a
∈
A

{\displaystyle a\in A}
and

b
∈
B

{\displaystyle b\in B}
. Then there is a path

f

{\displaystyle f}
from

a

{\displaystyle a}
to

b

{\displaystyle b}
, i.e.,

f
:
[
0
,
1
]
→
X

{\displaystyle f:[0,1]\rightarrow X}
is a continuous function with

f
(
0
)
=
a

{\displaystyle f(0)=a}
and

f
(
1
)
=
b

{\displaystyle f(1)=b}
. But then

f

−
1

(
A
)

{\displaystyle f^{-1}(A)}
and

f

−
1

(
B
)

{\displaystyle f^{-1}(B)}
are disjoint open sets in

[
0
,
1
]

{\displaystyle [0,1]}
, covering the unit interval. This contradicts the fact that the unit interval is connected.

== Exercises ==
Prove that the set

A
=
{
(
x
,
f
(
x
)
)

|

x
∈

R

}
⊂

R

2

{\displaystyle A=\{(x,f(x))|x\in \mathbb {R} \}\subset \mathbb {R} ^{2}}
, where

f
(
x
)
=

{

0

if

x
≤
0

sin
⁡
(

1
x

)

if

x
>
0

{\displaystyle f(x)=\left\{{\begin{array}{ll}0&{\text{if }}x\leq 0\\\sin({\frac {1}{x}})&{\text{if }}x>0\\\end{array}}\right.}
is connected but not path connected.