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 intervalunder a continuous function. Since the unit interval is connected, every interval inis connected.

If h is a continuous function with andthenis a path infromtoClearly 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 curveonDisconnect the sin curve into two open intervals A and B, withLet be the lower bound of B. Ifthen 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 coordinatewhile B includes the- coordinatewithA and B disconnect the curve fromtowhich 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 fromto the point Letbe the continuous function that describes the path fromtoLetbe the lower bound of the values offor whichForand by continuityRemove the segmentand reparameterize the path so thatfor 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 pointtowithandLetThencontains 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 andis compact, somust be compact,which is a contradiction. Hence the curve is not path connected.