# General Topology/Connected spaces

**Definition (connectedness)**:

Let be a topological space. is called **connected** if and only if whenever are two proper open subsets such that , then . A subset of a topological space is called **connected** if and only if it is connected with respect to the subspace topology.

**Proposition (characterisation of connectedness)**:

Let be a topological space. The following are equivalent:

- is connected
- Whenever are closed so that , then
- and are the only subsets of which are simultaneously open and closed

**Proof:** If is connected, suppose that are closed so that . Set and . Then , so that we find . Suppose then that is clopen (ie. open and closed), and . Then is the disjoint union of two nontrivial closed subsets, contradiction. Finally, if and , then and are both clopen.

**Example (two disjoint open balls in the real line are disconnected)**:

Consider the subspace of , equipped with the subspace topology. It is an example of a space which is not connected.

**Proposition (continuous image of a connected space is connected)**:

Let be a continuous function, and suppose that is connected. Then is connected with respect to its subspace topology (induced by ).

**Proof:** Suppose that and are two open subsets of such that and . By definition of the subspace topology, write and , where are open in . Since is continuous, and are open in . Further, , since any element in would be mapped to . Finally, every element in is either mapped to or to , so that , and is not connected, a contradiction.

**Example (the closed unit interval is connected)**:

Set with the topology induced by the Euclidean topology on . Then is connected.

**Proof:** Let be two open subsets of so that and . Suppose by renaming if necessary that . has an infimum, say . Since , pick by openness of an such that . Then , so that . Suppose that . Then for some , so that in particular , a contradiction to . Hence , but then pick so that and obtain that , a contradiction.

**Definition (connected component)**:

Let be a topological space and let . The set of all so that there exists which is connected and is called the **connected component** of .

**Proposition (topological spaces decompose into connected components)**:

Let be a topological space. Then

- ,

where the union is disjoint and each is the connected component of each of its points.

**Proof:** We prove that being contained within a common connected set is an equivalence relation, thereby proving that is partitioned into the equivalence classes with respect to that relation, thereby proving the claim. Indeed, it is certainly reflexive and symmetric. To prove it transitive, let and , where and are connected. We claim that is connected; once this is proven, and will lie in a common connected set (). Hence, let , where are open with respect to the subspace topology on , that is, , for suitable that are open in . Since is connected, or since and . Suppose, by renaming , if necessary, that , that is, . Note that by a similar argument, or , but is impossible, since then . Hence, and , so that , and is connected.

**Definition (path)**:

Let be a topological space. A **path** is a continuous function , where .

**Definition (concatenation of paths)**:

Let and are two paths such that . Then the **concatenation** of and is defined to be the path

- .

**Proposition (concatenation of paths is continuous)**:

Let and be two paths. Then is continuous.

**Proof:** We have

- and ,

both of which are continuous. We conclude since a function continuous when restricted to two closed subsets which cover the space is continuous.

**Definition (path-connectedness)**:

Let be a topological space. is called **path-connected** if and only if for every two points , there exists a path such that and . A subset is called **path-connected** iff, equipped with its subspace topology, it is a path-connected topological space.

That is, a space is path-connected if and only if between any two points, there is a path.

**Proposition (path-connectedness implies connectedness)**:

Let be a path-connected topological space. Then is also connected.

**Proof:** Suppose that , where are open and . Suppose there exist and , so that and are both proper nonempty subsets of . Then consider by path-connectedness a path such that and . is a connected subspace of since is a continuous image of the closed unit interval which is connected, and is then connected as the continuous image of a connected set, since the continuous image of a connected space is connected. On the other hand, and , where and are both open with respect to the subspace topology on , so that is not connected, a contradiction.

**Proposition (connectedness by path is equivalence relation)**:

Let be any topological space. Then the relation

is an equivalence relation.

**Proof:** For reflexivity, note that the constant function is always continuous. For symmetry, note that if we are given such that and , we may consider the path

- ,

which is continuous as the composition of continuous functions and has the property that and . Finally, whenever we have a path such that and , and another path such that and , then is a path such that and , so that transitivity holds.

**Definition (path-connected component)**:

Let be a topological space, and let be a point. The **path-connected component** of is the equivalence class of , where is partitioned by the equivalence relation of path-connectedness.

Here we have a partial converse to the fact that path-connectedness implies connectedness:

**Definition (local path-connectedness)**:

Let be a topological space. is called **locally path-connected** iff for every and every open set of such that , there exists an open neighbourhood of such that which is path-connected.

**Theorem (equivalence of connectedness and path-connectedness in locally path-connected spaces)**:

Let be a topological space which is locally path-connected. Then is connected if and only if it is path-connected.

**Proof:** First note that path-connected spaces are connected. Then suppose that is connected, fix , and define the set

- .

Note that is open, since if , then by local path-connectedness we may pick a path-connected open neighbourhood of , so that by applying concatenation, we see that all points in are in . On the other hand, is open, pretty much by the same argument: If and is a path-connected open neighbourhood of , then , since if would contain a point of , could be joined to by a path, concatenating a path from to to one from , in contradiction to . Hence is open and closed, and since , , so that by connectedness.

This theorem has an important application: It proves that manifolds are connected if and only if they are path-connected. Also, later in this book we'll get to know further classes of spaces that are locally path-connected, such as simplicial and CW complexes.

By substituting "connected" for "path-connected" in the above definition, we get:

**Definition (local connectedness)**:

Let be a topological space. is called **locally connected** if and only if for and every neighbourhood of , there exists a connected neighbourhood of such that .

## Exercises[edit | edit source]

- Prove that whenever is a connected topological space and is a topological space and is a continuous function, then is connected with the subspace topology induced on it by .
- Prove that similarly if is a path-connected top. space, top. space, continuous, then is path-connected with the subspace topology induced on it by .
- Prove that a topological space is connected if and only if, when for denotes the indicator function, the only indicator functions which are continuous are the ones where and ; here has the discrete topology.
- Let with the subspace topology induced by the Euclidean topology of . Prove that is not locally connected by proving that does not have a connected neighbourhood.