A connected topological space is a space that cannot be expressed as a union of two disjoint open subsets. Connectedness is a property that helps to classify and describe topological spaces; it is also an important assumption in many important applications, including the intermediate value theorem.
As with compactness, the formal definition of connectedness is not exactly the most intuitive. The more natural definition is that a space is connected if a curve can be drawn between any two points in the space ("connecting" the points). With the appropriate assumptions and definition, this down-to-earth definition becomes an important alternate notion of connectedness known as path-connectedness. Connected spaces are not always path-connected (although the converse is true).
Let be a topological space. Then is said to be disconnected if with open subsets of such that Otherwise, is connected.
An equivalent formulation of this definition involves clopen sets, defined as follows:
A clopen set is a set that is both closed and open.
Clearly, and are clopen, and the definition of connectedness is related to other clopen sets:
is connected if and only if the only clopen subsets of are and
This is clear, since if are open subsets with and then and are clopen. (This reasoning works in reverse as well.)
Let be a subset of a topological space Recall that inherits a topology from called the subspace topology, where the open sets of are the intersections of open sets of with Then a subset of is connected if and only if it is connected as a topological space, with the subspace topology.
Consider the subset Then is disconnected, because it is the union of and Neither of these are open subsets of but they are open subsets of with the subspace topology since they are the intersection of e.g. and with
Any interval in is connected.
All proofs of this result use some form of the completeness property of Here is one such proof. For simplicity's sake, we will only give the proof when the interval is closed, and without loss of generality the interval can be assumed to be in this case. The result for general intervals follows quickly from the result for closed intervals, by embedding a non-closed interval inside a closed interval with the same endpoints (left as an exercise for the reader).
So suppose with open, disjoint, nonempty. Without loss of generality, pick with Let (This is where completeness is used, to know that this supremum exists.)
If then find an interval contained entirely in this is possible since is open Note that is not in this interval because Since it follows that in this case This violates the description of as being larger than all elements of which are since it is not larger than e.g.
If then a similar argument shows that is not the least upper bound for the set So in both cases there is a contradiction.
Every topological space can be partitioned into connected components; these are simply the maximal connected subsets of ordered by inclusion. The connected components of are non-empty, disjoint, and their union is One way to see this is by considering the equivalence relation on points of defined by if and only if and are both contained in a connected set; then the connected components are the equivalence classes under this relation.
The fact that this relation is transitive requires the following important result:
Let be a topological space and two connected subspaces. Then if is non-empty, is connected.
Suppose is disconnected, e.g. where are disjoint, nonempty, and open in Then pick a point Suppose without loss of generality that So is nonempty, and is open in Now is open in and disjoint from and and cover Since is connected, the only possibility is that So and But now we can use that is connected: and are disjoint and nonempty and cover which is impossible. This contradiction shows that must be connected.
This section describes a more straightforward definition of a connectedness property, involving joining points by paths inside the topological space. First, it is necessary to define a path.
Let be a topological space. A path between two points is a continuous function such that and
A topological space is path-connected if there is a path between any two points in
A path-connected topological space is connected.
Suppose is path-connected but not connected. So with nonempty disjoint open sets. Take and let be the path between them. Then and are disjoint, nonempty, open subsets of which cover But this is impossible, since a closed interval in is connected.
The converse of this theorem is false. Here is one well-known counterexample in called the "topologist's sine curve."
Let be the subset of consisting of the points for all plus the origin Show that is connected but not path-connected.
Note that is path-connected, since it is the graph of a continuous function. Suppose is not connected; then an expression of as a disjoint union of and with leads to an expression of as a disjoint union of and The key point is that is a limit point of so is nonempty. This is a contradiction, since is path-connected and hence connected.
To see that is not path-connected, the idea is to show that any path starting at and staying inside is the trivial path which stays on The intuition is, roughly, that a path has no way to "jump" from the origin onto the rest of because is "infinitely wavy" near To make this formal, suppose is a path with Suppose it is not the trivial path. Let By continuity of Also by continuity, there is a such that the -coordinates of for are less than, say, Because is the infimum, there is some point in that interval for which But the interval must contain infinitely many points whose -coordinate is (the "infinitely wavy" part of the argument), which is a contradiction.
Continuous images: If is a continuous function, and is connected, then is connected.
This is a generalization of the argument that path-connected implies connected: if with nonempty, open, disjoint, then and are nonempty, open, disjoint sets whose union is which is impossible since is connected.
The same result holds for path-connected sets: the continuous image of a path-connected set is path-connected.
The proof is immediate: if is a path connecting to then is a path connecting to
A special case of this result is the intermediate value theorem: if is continuous and is any real number between and then there is a such that This is because the image of is path-connected, and hence connected. If then the sets and are open, disjoint, nonempty sets whose union is which is a contradiction.
Closure: The closure of a connected set is connected in fact, any set such that is connected This is a generalization of the topologist's sine curve argument: adding limit points does not affect connectedness.
Unions and intersections: The union of two connected sets is connected if their intersection is nonempty, as proved above. But if their intersection is empty, the union may not be connected e.g. two disjoint open intervals in
The intersection of two connected sets is not always connected.
Show that no two of the sets and are homeomorphic.
Deleting any point from gives a disconnected space. There is exactly one point in which can be deleted to give a connected space. There are exactly two points in which can be deleted to give a connected space. If there were a homeomorphism between or and then the image of would be with one (or two) points deleted, which contradicts the fact that continuous images of connected sets are connected. Similarly, if there were a homeomorphism the image of would be with two points deleted, which cannot be connected.
- Stella, B. Union et intersection d'ensembles. Retrieved October 30, 2006, from https://en.wikipedia.org/wiki/Connected_space#/media/File:Union_et_intersection_d%27ensembles.svg