A topological space
is said to be path connected if for any two points
there exists a continuous function
such that
and
- All convex sets in a vector space are connected because one could just use the segment connecting them, which is
.
- The unit square defined by the vertices
is path connected. Given two points
the points are connected by the function
for
.
The preceding example works in any convex space (it is in fact almost the definition of a convex space).
Let
be a topological space and let
. Consider two continuous functions
such that
,
and
. Then the function defined by
Is a continuous path from
to
. Thus, a path from
to
and a path from
to
can be adjoined together to form a path from
to
.
Each path connected space
is also connected. This can be seen as follows:
Assume that
is not connected. Then
is the disjoint union of two open sets
and
. Let
and
. Then there is a path
from
to
, i.e.,
is a continuous function with
and
. But then
and
are disjoint open sets in
, covering the unit interval. This contradicts the fact that the unit interval is connected.
- Prove that the set
, where ![{\displaystyle f(x)=\left\{{\begin{array}{ll}0&{\text{if }}x\leq 0\\\sin({\frac {1}{x}})&{\text{if }}x>0\\\end{array}}\right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8506a3ca2a83a59f725a45a7613c3e0fc0abc93e)
is connected but not path connected.