## 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 in is connected.

If h is a continuous function with and then is a path in from 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 to which 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 from to Let be the lower bound of the values of for 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 point to with and Let Then contains at most one point of B. I want to show Closure of contains all of B so is not closed, and therefore not compact. But f is continuous and is compact, so must be compact,which is a contradiction. Hence the curve is not path connected. 