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}}
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}
.
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.
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.