Real Analysis/Connected Sets

Intuitively, the concept of connectedness is a way to describe whether sets are "all in one piece" or composed of "separate pieces". For motivation of the definition, any interval in should be connected, but a set consisting of two disjoint closed intervals and should not be connected.

Definition A set in in is connected if it is not a subset of the disjoint union of two open sets, both of which it intersects.
Alternative Definition A set is called disconnected if there exists a continuous, surjective function , such a function is called a disconnection. If no such function exists then we say is connected.
Examples The set cannot be covered by two open, disjoint intervals; for example, the open sets and do not cover because the point is not in their union. Thus is connected.
However, the set can be covered by the union of and , so is not connected.

Path-Connected edit

A similar concept is path-connectedness.

Definition A set is path-connected if any two points can be connected with a path without exiting the set.

A useful example is  . Any two points a and b can be connected by simply drawing a path that goes around the origin instead of right through it; thus this set is path-connected. However,   is not path-connected, because for   and  , there is no path to connect a and b without going through  .

As should be obvious at this point, in the real line regular connectedness and path-connectedness are equivalent; however, this does not hold true for   with  . When this does not hold, path-connectivity implies connectivity; that is, every path-connected set is connected.

Simply Connected edit

Another important topic related to connectedness is that of a simply connected set. This is an even stronger condition that path-connected.

Definition A set   is simply-connected if any loop completely contained in   can be shrunk down to a point without leaving  .

An example of a Simply-Connected set is any open ball in  . However, the previous path-connected set   is not simply connected, because for any loop p around the origin, if we shrink p down to a single point we have to leave the set at  .