The Difference Between Connectedness and Path Connectedness
A space is connected if it cannot be partitioned into at least two non empty disjoint open sets, or equivalently, the only sets that are both open and closed are the entire space and the empty set. Since connectedness is a topological property preserved by continuous functions, an space is connected if there is no continuous function from the space onto the set
A path is the image of the closed interval
under a continuous function. Since the unit interval is connected, every interval inis connected.
If h is a continuous function with
and
then
is a path infrom
to
Clearly x is connected to itself, and path connectivity is symmetric and transitive, so path connectedness is an equivalence relation. All the points in a path connected component can be connected to each other. A set, or space, is path connected if it consists of one path connected component.
The continuous image of a path is another path, by the property of composition of continuous functions so the image of a path connected component is also path connected. Two topological spaces are homeomorphic if there is a continuous function from one to the other. Homeomorphic spaces have the same number of path connected components.
Suppose a path connected component C is not connected. Separate C into two disjoint open sets and draw a path from a point in one set to a point in the other. The path is separated so that each part is a subset of disjoint open sets, contradicting the fact the connectedness of the path. Therefore path connected implies connected.
Connected and path connected are not equivalent, as shown by the curve
on
Disconnect the sin curve into two open intervals A and B, with
Let
be the lower bound of B. If
then every open ball about this point includes points from A and B, so if 
belongs to either A or B, that set is not open. This is a contradiction, hence
B includes coordinates arbitrarily close to 0. A is open, so A includes a positive coordinate. Let A include the x coordinate
while B includes the- coordinate
with
A and B disconnect the curve from
towhich is homeomorphic to a closed interval. A closed interval cannot be disconnected, hence the curve is connected.
Next suppose the set is path connected, so there is a path from
to the point
Let
be the continuous function that describes the path fromto
Letbe the lower bound of the values offor which
For
and by continuity
Remove the segment
and reparameterize the path so that
for values of arbitrarily close to 0.
Now show the curve is not path connected. Assume to the contrary that there exists a path
from the pointto
with
and
Let
Then
contains at most one point of B. I want to show Closure ofcontains all of B sois not closed, and therefore not compact. But f is continuous and
is compact, somust be compact,which is a contradiction. Hence the curve is not path connected.
