## Locally Connected Spaces

Recall that each topological space *X* is the set-theoretic disjoint union of its connected components, but in general (e.g. for *X*=**Q**) fails to be the topological disjoint union. The problem is that the connected components in general aren’t open in *X*. We’ll seek to right this wrong here, by looking at a specific class of topological spaces.

Definition. A topological space X is said to belocally connectedif

- for each open subset and there exists a connected open subset V of X such that

Note that since *U* is open in *X*, any subset is open in *U* if and only if it’s open in *X*. In such instances, we’ll sometimes abuse our language and say “*V* is open” without specifying the ambient space.

One useful way to judge if a space is locally connected is as follows.

Proposition 1. A topological space X is locally connected if and only if it has a basis all of whose elements are connected.

**Proof**.

(→) Let be open; we need to show it’s a union of *connected* open subsets of *X*. Indeed, for by local connectedness, there’s a connected open subset containing *x*. So *U* is indeed a union of connected open subsets.

(←) Let be open and Now we can write *U* as a union of connected open subsets. One of these (say *V*) must contain *x*. ♦

The key property we wish to prove is:

Theorem 2. If X is locally connected, then every connected component of X is open in X. Hence X is the topological disjoint union of its connected components.

**Proof**.

Let , where *Y* is a connected component of *X*. By definition, *x* is contained in some open connected subset *U* of *X*. Since *Y* is a maximal connected set containing *x*, we have This shows that *Y* is open in *X*. ♦

**Important Non-Example**

Take our favourite topologist’s sine curve where:

and

We saw that *X* is connected. However it is *not locally-connected* since for any , the ε-neighbourhood of the origin contains more than one (in fact infinitely many!) connected components:

This also shows that **connected spaces are generally not locally connected**.

Corollary. A locally connected space X is totally disconnected if and only if it’s discrete.

**Proof**.

If *X* is totally disconnected, each {*x*} is a connected component; since *X* is locally connected, each connected component is open in *X*. ♦

## Properties of Locally Connected Spaces

Similar to the case of locally compact, the following result explains why the property is “local”.

Proposition 3. Let X be a topological space.

- If is open, and X is locally connected, then so is U.
- If is a union of open subsets and each U
_{i}is locally connected, then so is X.

**Proof**.

1st statement: let where *U’* is open in *U*. Then *U’* is open in *X* so there exists a connected open subset *V* of *X* such that This *V* is also open in *U*.

2nd statement: let be open and . Then for some *i*. By local connectedness of *U _{i}*, and there exists a

*connected*open subset

*V*of

*U*∩

_{i}*U*such that Thus, ♦

It’s not true that if *f* : *X* → *Y* is continuous and *X* is locally connected, then so is *f*(*X*). Indeed, let *X* = **Q** with the discrete topology and *Y* = **Q** as a subspace of **R**. The identity map on the underlying set **Q** then gives a surjective continuous map but *Y* is not locally connected. [ Recall that if we replace “locally connected” with “connected”, then this is true. ]

Closed subsets of locally connected spaces are not locally connected in general. E.g. take *X* = **R** and Then *Y* is closed in *X*, but there’s no connected open subset of 0 in *Y*.

–

Proposition 4.If X and Y are locally connected topological spaces, then so is X × Y.

**Proof**.

Since *X* is locally connected, there is a basis *B _{X}* comprising of open connected subsets. Likewise, pick such a basis

*B*for

_{Y}*Y*. Then is a basis of

*X*×

*Y*comprising of open connected subsets. ♦

*This does not hold for infinite products*. For example, let *X* = {0, 1}^{N}, where {0, 1} is given the discrete topology. We claim that *X* is totally disconnected.

Indeed, suppose some connected component *Y* contains where for some *n*. Then projecting to the *n*-th component gives a surjective map which violates the theorem that the continuous image of a connected set is connected.

Hence, *X* is a totally disconnected space which is not discrete (since it’s compact by Tychonoff theorem), so it can’t be locally connected (see corollary to theorem 2).

**Examples**

- Any discrete set is locally connected since we can take
*V*={*x*}. - Since the open intervals in
**R**are connected,**R**has a basis of connected open subsets and is thus locally connected. - Hence
**R**^{n}is also locally connected by proposition 4, as is any open subset of**R**^{n}by proposition 3. - Let
*X*= [0, 1]; then*X*is connected and locally connected. For example, at 0, any open subset must contain for some ε>0, which is open in*X*. **Q**is totally disconnected yet not discrete, so it’s not locally connected.

## Locally Path-Connected Spaces

Correspondingly, we have locally path-connected spaces.

Definition. A topological space X islocally path-connectedif

- for each open subset and there exists a path-connected open subset V of X such that

The earlier properties all carry over since the proofs can be replicated by replacing “path-connected” with “connected” whenever it appears in the text.

Proposition 5. A topological space X is locally path-connected if and only if it has a basis all of whose elements are connected.

Theorem 6. If X is locally path-connected, then every path-connected component of X is open in X. Hence X is the topological disjoint union of its path-connected components.

Proposition 7. Let X be a topological space.

- If is open, and X is locally path-connected, then so is U.
- If is a union of open subsets and each U
_{i}is locally path-connected, then so is X.

Proposition 8. If X and Y are locally path-connected, then so is X × Y.

**Examples (same as earlier).**

- Any discrete set is locally path-connected since we can take
*V*={*x*}. - Since the open intervals in
**R**are path-connected,**R**has a basis of path-connected open subsets and is thus locally path-connected. - Hence
**R**^{n}is also locally path-connected, as is any open subset of**R**^{n}. - Let
*X*= [0, 1]; then*X*is path-connected and locally path-connected. For example, at 0, any open subset must contain for some ε>0, which is open in*X*. **Q**is totally disconnected; since each connected component is a disjoint union of path components, we see that the only path components of**Q**are {*x*}. Hence,**Q**is*not locally path-connected*, for if it were,**Q**would have to be the disjoint union of all {*x*} and hence discrete.

Let’s see if we can find counter-examples similar to those for local connectedness.

## Non-Properties

**1. Not true: if X is path-connected, then it’s locally path-connected.**

Take the circle and consider an infinite sequence of wheel spokes: and:

for *n* = 1, 2, … .

Now take :

Now there’s no path-connected open subset *V* such that On the other hand, *X* is clearly path-connected.

**2. Not true: if f : X → Y is continuous and X is locally path-connected, then so is Y.**

Same example as before: let *X*=**Q** with the discrete topology and *Y*=**Q** as a subspace of **R**. Then the identity map on **Q** is continuous and *X** *is locally path-connected, but *f*(*X*) = *Y* is not.

**3. Not true: a product of infinitely many locally path-connected spaces is locally path-connected.**

Same example as before: *X* = {0, 1}^{N}, where {0, 1} is given the discrete topology. We saw earlier that *X* is totally disconnected, so the components (and hence path components) are singleton sets. On the other hand, *X* is not discrete, so it cannot be locally path-connected.

## Relationship Between (Locally) Connected & Path-connected

Finally, let’s examine the relationship between the two notions. Since path-connected sets are connected, we have:

**1. A locally path-connected space is also locally connected.**

The converse isn’t true.

**2. A locally connected space is not locally path-connected in general.**

This is hard: one can find a counter-example in *Munkres*, “*Topology*“, 2nd edition, page 162, chapter 25, exercise 3.

**3. If X is connected and locally path-connected, then it’s path-connected.**

Pick any path component *Y* of *X*. Since *X* is locally path-connected, *Y* is open in *X*. The complement *X*–*Y* is a union of path components, each open in *X*, so it’s open too. Thus *Y* is a clopen subset of *X* and we must have *Y*=*X*.

## Summary

The following summarises various properties of compactness, connectedness and path-connectedness. Note: for any property that mentions local compactness, we also assume *X* is Hausdorff.

C | Compact | Connected | Path-connected |

If X is C, and f:X→Y is continuous, then f(X) is C. |
🙂 | 🙂 | 🙂 |

If each X is C, then their product also is C._{i} |
🙂 | 🙂 | 🙂 |

C implies (locally C). | 🙂 | – | – |

If X is locally C, then so is any open subset U of X. |
🙂 | 🙂 | 🙂 |

If X is locally C, then so is any closed subset of X. |
🙂 | – | – |

If is a union of open subsets and each U is locally C, then so is _{i}X. |
🙂 | 🙂 | 🙂 |

If f:X→Y is continuous and X is locally C, then so is f(X). |
– | – | – |

If X and Y are locally C, then so is X × Y. |
🙂 | 🙂 | 🙂 |

If each X is locally C for infinitely many _{i}i, then so is their product. |
– | – | – |

What do you mean by “Recall that each topological space X is the set-theoretic disjoint union of its connected components, but in general (e.g. for X=Q) fails to be the topological disjoint union. “?

Isn’t Q also disjoint union of connected components?

Sorry for the late reply. Q is set-theoretically the disjoint union of its connected components, but not the topological disjoint union, simply because a singleton point is not an open subset of Q.

You state that $Y$ and $Z$ are connected components of topologist’s sine curve, a connected space. However, any connected space should have itself as the only connected component.

You’re right! Corrected it.