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Connectedness and Paths

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tagssimply-connectedness, path-connectedness, connected-components, local-connectedness

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Connectedness

Connectedness

A topological space is called connected if there do not exist two disjoint nonempty open sets and such that .

e.g.

  • The set of rational numbers in the standard Euclidean topology is not connected because with and . Note that this also means that a subspace of a connected space need not be connected.
  • Let be endowed with the standard Euclidean topology. A set is connected if and only if is an interval.

Theorem

Let and be topological spaces. If is continuous and is connected, then is connected, i.e. the continuous image of a connected set is connected.

Proof We prove the contrapositive. Suppose is not connected, then there exist disjoint nonempty open sets and in such that . Then and are disjoint nonempty open sets in with which shows that is not connected.

Corollary

Every continuous map from a connected topological space to a discrete topological space is constant.

Proof This is because the only connected subsets of a discrete space are the singletons.

Intermediate Value Theorem

Let be a connected topological space and a continuous function, where is endowed with the standard topology. If takes the values and , then takes all the values between and .

Proof By the given conditions, is connected and hence it is an interval. Since , any value between and must be in .

Lemma

If is connected and , then is connected. In particular, the closure of a connected set is connected.

Proof Since , we can write for for some where is the set of limit points of . Without loss of generality, we assume that , so . Suppose for nonempty open sets . Since is connected, either or . Without loss of generality, assume . Then is nonemty. Pick some , then is an open neighborhood of , so it must intersect , yielding a contradiction. Therefore, is connected.

Proposition

A topological space is connected iff the only both open and closed sets are the empty set and itself.

Proof Suppose is both open and closed. Then and are disjoint open sets with , which implies that either or is empty.

Path Connectedness

Path

Let be a topological space. A path in joining two points is a continuous function such that and .
If the start point and end point coincide, we call it a loop.

Path Connectedness

A topological space is called path connected if for every pair of points can be joined by a path in .

Theorem

A path connected topological space is connected. In general, connected space is not necessarily path-connected.

Proof If is not connected, then there exist disjoint nonempty open sets , in with . Let and . Since is path-connected, there is a path joining to . By the continuity of , and are disjoint nonempty open sets in with Therefore is not connected, which is a contradiction.

e.g.

  • Any convex set in a topological vector space in is path connected;
  • is path connected for ;
  • The topologist’s sine curve is connected but not path connected:

Theorem

Continuous image of a path-connected space is also path-connected.

Proof This is clear because of the following diagram:

Proposition

If is an open set in , then is connected if and only if is path connected.

Proof It suffices to show that if is nonempty connected, then it is path connected. Fix some , we define . Clearly because . By the proposition, it is enough to show that is both open and closed in , which implies that , hence is path connected. For any , since is open, there is a open ball contained in . Let be the path from to . Note that for all , we can define a path from to by connecting and , so , which shows that is open in .
connectedness_in_R^n
Now we show that is closed in . Let be a sequence converging to . As is open, there is some open ball , and for all sufficiently large . Similarly by concatenation of paths, we can define a path joining to and then to , hence . This shows that is closed in . Therefore, is path connected.

Proposition

The following holds:

  1. Every quotient of a (path) connected space is (path) connected;
  2. The union of a family of (path) connected subspaces of that have a point in common is (path) connected;
  3. Any product of (path) connected spaces is (path) connected.

Proof (1) is immediate from the theorem; For (2), path connectedness is clear, and we will only prove that it is connectedness here. Suppose is a family of connected subspaces of with a point . Suppose separates , without loss of generality, we assume that , then for all , which implies that , and .
For (3), path connectedness is clear by the universal property, and we will only prove connectedness. We will first show that the product of two thus any finite connected spaces is connected, and then extend the result to the infinite case. In fact, fix some , we have so by (2), is connected if both and are connected.

Components

Components

A (path-) component of a topological space is a maximal (path-) connected subspace of .

Proposition

The following holds for a topological space :

  1. The (path) components form a partition of ;
  2. Any (path) connected subset is contained in a unique (path) component;
  3. Each path component is contained in a single component and each component is a disjoint union of path components.

Proof By (2) of the proposition, we know that whenever (path) components and share a point, then is also (path) connected, so by maximality, the components must be disjoint; Additionally, the components cover because every singleton is connected. (2) is obvious. (3) is because each component is a topological space and hence partitioned into path components.

Proposition

Components of a topological space are always closed.

Proof This is a direct consequence of the lemma, because the closure of a component is also connected, so the maximality implies that the component itself has to be closed.

Remark

Note that in general components need not be open. For example, the components of (with the subspace topology from ) are singleton sets, which are not open.
More importantly, path components need not be open nor closed. Think about the topologist’s sine curve, one of the path component is open, and the other is closed.

Locally (Path) Connectedness

Sometimes being (path) connected globally is not enough. We can still have “bad” and “wild” spaces that are separated locally. We have the following variation notion on connectedness:

Locally (Path) Connected

A topological space is locally (path)-connected if for any point and any open neighborhood of , there exists a (path) connected open neighborhood containing .
Equivalently, this means admits a basis of (path) connected open sets.

In general, there are no implications between the local and global versions. Here are some examples:

e.g.

  • Consider . Clearly it is path connected but not locally path-connected because for any point , any open neighborhood of contains points of the form for , which cannot be connected by a path in ;
  • is locally connected but not connected;
  • The topologist’s sine curve is connected but not locally connected.

Similar to the global version, local path-connectedness is also stronger than local connectedness:

Proposition

If is locally path connected, then it is locally connected.

Proof Immediate from the basis characterization of local (path-) connectedness and the fact that every path connected space is connected.

Proposition

If is locally (path-) connected, then every open subset is locally (path-) connected. Moreover, every (path-) component is open.

Proof If is locally (path-) connected, then it has a (path-) connected basis, then the every basis element that is included in an open subset of is also a (path-) connected basis element for .
We now prove that every component is open. Suppose is a (path-)component. Then for any , there is a (path-) connected open set containing , and because is maximal. Hence is open.

The following proposition and corollary give a sufficient condition for path-connectedness being the same as connectedness. It can explain the equivalence of path-connectedness and connectedness on $\R^{n}$ in a clean way.

Proposition

Suppose is locally path-connected, then the path components of coincide with the components of .

Proof Since every component is a partition of path components, it suffices to show that every component is path-connected. Suppose is a component, then it is a partition of path components, and each path component is open by the previous proposition. If has two distinct path components and , then they will separate , which is a contradiction. Hence has to be path-connected.

Corollary

If a topological space is locally path-connected, then it is connected iff it is path-connected.

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