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A topological space is said to be **locally connected** at *x* (where *x* is a point of the space to be called *X*), if for each neighbourhood *V* of *x*, there is a connected neighbourhood *U* of *x* contained in *V*. If a space is locally connected at each of its points, we call the space locally connected. Suprisingly, local connectedness and connectedness are independent properties; a space may satisfy one of them without satisfying the other (see the examples below).

## ExamplesEdit

1. If a space has a basis consisting entirely of connected sets, then it is locally connected (to see this, let *x* be a point of the space. Then for each neighbourhood *V* of *x*, there is a basis element *B* containing *x* which is connected and contained in *V*. Then *B* is the desired connected neighbourhood of *x* contained in *V*).

2. From example 1, we can conclude that the set of real numbers **R** is locally connected and connected (since it is a linear continuum).

3. The set of rational numbers **Q** is neither locally connected nor connected. The same is true for **R** in the lower limit topology. In fact, a totally disconnected space cannot be locally connected unless it has the discrete topology.

4. The finite union of intervals is locally connected. In other words, elementary sets (see measure theory) in **R** are locally connected. In particular, the set

*A*= (-1,0) ∪ (0,1)

is locally connected but not connected.

5.The topologist's sine curve (if *f*(*x*) = sin (1/*x*), then the topologist's sine curve is just the closure of the image of (0,1] under *f*), is an example of a connected space which is not locally connected. The fact that the topologist's sine curve is connected follows from:

- a) The set
*S*=*f*((0,1]) is connected since it is the image of a connected space under a continuous map. - b) The closure of a connected space is connected.

The space is not locally connected at any point in the set *B* = [Closure (*S*)] – *S*. It is locally connected at every other point though.

## Local path connectednessEdit

A topological space, *X*, is locally path connected, if for each point x, and each neighborhood *V* of *x*, there is a path connected neighbourhood *U* of *x* contained in *V*. Similar examples to the previous ones, show that path connectedness and local path connectedness are independent properties. We will give a few more examples.

- Examples

1. A locally path connected space is always locally connected. The proof uses the fact that every path connected space is connected.

2. An example of a path connected space that is not locally path connected is the comb space: if *K* = {1/*n* |*n* is a natural number}, then the comb space is defined by:

*C*= (*K*× [0,1]) ∪ ({0} × [0,1]) ∪ ([0,1] × {0}]

The comb space is path connected (this is trivial) but locally path connected at no point in the set *A* = {0} × (0,1]. It is however locally path connected at every other point.

## Weakly Locally ConnectedEdit

A weaker property that a topological space can satisfy at a point is known as ‘weakly locally connected’:

**Definition**

Let X be a topological space and x a point of X. If for every neighbourhood V of x, there is a connected subspace A of V containing a neighbourhood U of x, the topological space is said to be weakly locally connected at x. If the space is weakly locally connected at each of its points, it is said simply to be weakly locally connected.

We will give a few examples and there will be one important theorem on this concept later.

**Examples**

1. Every locally connected space is trivially weakly locally connected (let A = U in the definition).

2. Note that neither the comb space nor the topologist’s sine curve is weakly locally connected. It is easy to see that these two spaces are not weakly locally connected at each of the points that they aren’t locally connected.

3. A totally disconnected space cannot be weakly locally connected unless it has the discrete topology (in other words, a totally disconnected space is weakly locally connected if and only if it has the discrete topology). An example is R in the lower limit topology which doesn’t have the discrete topology and hence cannot be weakly locally connected at any point.

4. We will prove shortly that if a topological space is weakly locally connected (that is weakly locally connected at EVERY point), then it is locally connected at every point. However, the proof of this fact requires another notion related to local connectedness which is to be discussed in the next section.

5. The previous property may seem strange. If every weakly locally connected space is locally connected, then why call a space ‘weakly’ locally connected? It doesn’t seem to be a weaker condition. However, if a space is weakly locally connected at just one point, then it need not be locally connected at that point. An example of such a space is the broom space. For this space to be locally connected at that point, it needs to be weakly locally connected in a NEIGHBOURHOOD of that point. Hence the definition of a weakly locally connected space does indeed make sense.

## Components and path componentsEdit

Define an equivalence relation ~ on a topological space *X*, so that we declare x~y if there is a connected subspace *A* of *X* containing both *x* and *y*. The collection of all equivalence classes are called the components of X (this is indeed an equivalence relation as the reader can check).

Define another equivalence relation ~_{2} (different from the first) to set *x*~_{2}*y* if there is a path from *x* to *y*. Equivalently, *x*~_{2}*y* if and only if there is a path connected subspace of *X* containing *x* and *y*. The collection of all equivalence classes with this equivalence relation are called the path components of the space (this is indeed an equivalence relation as the reader can check).

Being equivalence classes, the components are disjoint, nonempty subsets of the topological space *X*, whose union is X. The same is true for the path components.

- Properties

1. Intuitively, the components are the largest possible connected subsets of *X* and the path components are the largest possible path connected subsets of *X*.

2. The components are always connected (to see this let *C* be a component (the equivalence class containing a certain point *x* of the space). Then *C* is nonempty since *x*~*x*. For each *y* in *C*, let *A*_{y} be a connected subset of *X* containing both *y* and *x* (since *y*~*x*). Take the union of all such *A*_{y} for *y* in *C*. This union is *C*. Since this is the union of connected spaces having the point *x* in common, this union is connected. Therefore, *C* is connected). A similar argument shows that the path components are always path connected and hence connected.

3. Every connected subset of *X* lies in a component of *X*. This justifies property 1. To see this, let *A* be connected (and nonempty or else the proof is trivial), and let *x* belong to *A*. Then if *y* belongs to *A*, *A* is a connected subset of *X* containing both *y* and *x* so that *y*~*x*. Since *y* was arbitrary, we have shown that *A* is a subset of the component containing x. A similar proof shows that every path connected subset of *X* is a subset of a path component of *X*.

4. The components are always closed. To see this we note that if *C* is a component, then *C* is connected, so that the closure of *C* is connected. Now, C is always a subset of the closure of *C* so that it is sufficient to verify that the closure of *C* is a subset of *C*. But we know that the closure of *C* is connected, and every connected subset of *X* is the subset of exactly one component. Therefore, *C* = Closure(*C*) and *C* is closed. This proof fails for the path components since the closure of a path connected space need not be path connected (for example, the topologist's sine curve).

5. If there are only finitely many components, then the components are also open. If *C* is a component, then its complement is the finite union of components and hence closed. Therefore, *C* is open. This proof fails for infinitely many components, as the set of rational numbers show.

- Examples

1. The set of all real numbers has precisely one component; the whole space. In fact every connected space has only one component.

2. The set:

*A*= (-∞,0)∪(0,+∞)

has precisely two components; (-∞, 0) and (0, +∞). Note how these are the largest possible connected subsets of this space.

3. The set of rational numbers has countably many components. Each one point set is a component. In fact, for any totally disconnected space, the one point sets are the components of the space.

4. The path components of the spaces in examples 1,2 and 3 are exactly the same as the components. However, they are not the same in general. An example of this is the topologist's sine curve which has one component (since it is connected), but has two path components:

*A*= {0} × [-1,1]

*B*=*f*((0,1]) where*f*is the function defined by*f*(*x*) = sin(1/*x*)

Notice that *A* is closed; but not open. Also, *B* is open; but not closed. This shows that even though the components are always closed, the same thing does not hold for the path components.

5. In general, the path components are subsets of the components. To see this we note that the path components of a topological space are path connected and hence connected. Since connected subsets of *X* lie in a component of *X*, the result follows. We will prove later that the path components and components are equal provided that *X* is locally path connected.

6. The set *I* × *I* (where *I* = [0,1]) in the dictionary order topology has exactly one component (because it is connected) but has uncountably many path components. Indeed, any set of the form {*a*} × *I* is a path component for each *a* belonging to *I*.

7. Let *f* be a continuous map from **R** to **R**_{ℓ} (**R** in the lower limit topology). Since **R** is connected, and the image of a connected space under a continuous map must be connected, the image of **R** under *f* must be connected. Therefore, the image of **R** under *f* must be a subset of a component of **R**_{ℓ}. Since this image is nonempty, the only continuous maps from **R** to **R**_{ℓ}, are the constant maps.

## QuasicomponentsEdit

The concept of a quasicomponent is similar to that of a component and only brief attention will be given to it. We first make a definition:

**Definition**

Let X be a topological space. Define an equivalence relation on X to declare x~y if there is no separation of X into sets A and B such that x is an element of A and y is an element of B.

We note that the relation given in the definition is indeed an equivalence relation as the reader can readily check.

**Properties**

1. The components are always contained in the quasicomponents of X. Before proving this we will let x~_{1}y if there is a connected subset of X containing both x and y (which is the equivalence relation used in defining components), and x~_{2}y if x~y as in the definition given in this section. If x~_{1}y, then there is a connected subset, C, of X containing x and y. If there was a separation of X into sets A and B such that x is an element of A and y is an element of B, then C would intersect both A and B. This implies in particular that ‘C intersection A’ and ‘C intersection B’ would form a separation on C contradicting the connectedness of C. Therefore, we must have that x~_{2}y.

2. From the previous property, we may conclude that the path components of a space always lie in the quasicomponents of a space (since the path components always lie in the components).

**Examples**

1. There is only one quasicomponent of a connected space; the space itself. This follows from property 1.

2. In the next section, we will prove that if X is a locally connected topological space, then the quasicomponents of X are equal to the components of X.

3. If X is locally path connected, it is locally connected (as mentioned earlier) so that the components of X are equal to the quasicomponents of X. Since if X is locally path connected, the components of X and the path components of X are equal (which we shall prove in the next section), it follows that the quasicomponents, path components, and components are all equal if X is a locally path connected space.

4. Property 2 still holds if X is weakly locally connected (that is, weakly locally connected at each of its points). This is because a weakly locally connected space is always locally connected (which we shall prove imminently).

## TheoremsEdit

In this section, we will prove theorems relevant to the material in this article.

- Theorem 1

A topological space is locally connected if and only if the components of open sets are open.

- Proof

Suppose *X* is locally connected (where *X* is the topological space in question). Let *V* be open in *X* and let *C* be a component of *V*. For each *x* in *C*, choose a connected neigbourhood *U* of *x* contained in *V* (by the local connectedness of *X*). Since *U* is connected, *U* is a subset of *C*. It follows that *C* is the union of open sets and is thus open.

Conversely, suppose that the components of open sets in a topological space, *X*, are open. Let *x* belong to *X* and let *V* be a neigbourhood of *x*. Let *C* be a component of *V* containing *x*. Then *C* is open by hypothesis. Therefore, *C* is a connected neigbourhood of *X* contained in *V* so that *X* is locally connected at *x*. Since *x* was arbitary, *X* is locally connected.

- Theorem 2

A topological space is locally path connected if the path components of open sets are open.

- Proof

The proof is similar to theorem 1 and is omitted.

- Theorem 3

The components and path components of a topological space, *X*, are equal if *X* is locally path connected.

- Proof

Let *P* be a path component of *X* containing *x* and let *C* be a component of *X* containing *x*. We know that *P* is a subset of *C* by example 5 in the previous section. Suppose *P* is a proper subset of *C*. Since *X* is open, *P* is open in *X* by theorem 2. Take the union of all those path components of *X* disjoint from *P* and intersect them with *C*; call the union of all the resulting sets *Q*. Then, *P* and *Q* form a separation on *C*, since they are disjoint, nonempty (*P* is nonempty since *x* is in *P*; *Q* is nonempty, because we are assuming that *P* is a proper subset of *X*), and open in *C* (*Q* is open in *C* by theorem 2 and the definition of the subspace topology). This contradicts the connectedness of *C*.

- Theorem 4

Let *X* be a locally path connected space. Then every open, connected subset of *X* is path connected.

- Proof

Let *U* be an open connected subset of *X* and let *x* belong to *U*. Let *A* be the set of all points in *U* that can be joined by a path to *X*. Then *A* is open. If *y* is in *A*, then there is a path connected subset *V* of *U* containing *y* (by the local path connectedness of the space). Then we assert that *V* is a subset of *A*. If *z* is in *V*, then there is a path from *z* to *y*. Since there is a path from *x* to *y* (because *y* belongs to *A*), we can ‘paste’ these paths together to form a path from *x* to *z*. Therefore, *z* is in *A*. Therefore, *V* is a subset of *A* and *A* is open. Now if *y* is in *U*−*A*, then there is no path from *y* to *x*. If *V* is a path connected neigbourhood of *y* contained in *U*, then *V* is a subset of *U*−*A*. If *z* is in *V*, and if there is a path from *z* to *x*, then there would be a path from *x* to *y* since there is already a path from *z* to *y*. This cannot be since *y* is in *U*−*A*. Therefore, there is no path from *z* to *x* and *z* is in *U*−*A*. It follows that *V* is a subset of *U*−*A* and *U*−*A* is open. We know already that *A* is nonempty (*x* is in *A*). If *U*−*A* is nonempty, then *A* and *U*−*A* will form a separation on *U* contradicting the connectedness of *U*. Therefore, every point in *U* can be joined by a path to *x*. From this it follows that *U* is path connected.

**Theorem 5**

Let X be a weakly locally connected space. Then X is locally connected.

Proof

From theorem 1, it is sufficient to show that the components of open sets is open. Let U be open in X and let C be a component of U. Let x be an element of C. Then x is an element of U so that there is a connected subspace A of X contained in U and containing a neighbourhood V of x. Since A is connected and A contains x, A must be a subset of C (the component containing x). Therefore, the neighbourhood V of x is a subset of C. Since x was arbitrary, we have shown that each x in C has a neighbourhood V contained in C. This shows that C is open relative to U. Therefore, X is locally connected.

**Theorem 6**

Let X be a topological space. If X is locally connected, the quasicomponents of X are equal to the components of X.

Proof

We noted earlier in the section on quasicomponents that each component of the space is contained in a quasicomponent of the space. It suffices to show that the quasicomponents always lie in the components if the space is locally connected. From theorem 1, it follows that the components of X are open (since X is locally connected). Therefore, if x~_{1}y doesn’t hold (x is not equivalent to y under the equivalence relation defining components), then the component containing x is both open and closed (components are always closed; the component containing x is open because X is a locally connected topological space), so that there is a separation of X into sets A and B such that A contains x and B contains y (A is the component containing x, and B = X-A). Therefore, x cannot be equivalent to y under the equivalence relation defining quasicomponents. Therefore, we have proved that if y is not in the component containing x, y cannot be in the quasicomponent containing x. From elementary set theory, this implies that if y is in the quasicomponent containing x, it must be in the component containing x. Therefore, the quasicomponents of X are equal to the components of X.

- Some applications of the theorems

1. From theorem 4, we can conclude that every open connected subset of **R** is path connected (since **R** is locally path connected). Also, every open, connected subset of **R**^{2} is path connected.

2. The set of rational numbers **Q** is not locally connected since the components of **Q** are not open in **Q** (see theorem 1).

3. The components and path components of an elementary subset of R are the same. Also, the elementary subsets of **R** are the finite union of intervals, since every elementary set is locally path connected.

4. Take the point, p = (0.5,0.5) in the plane. For each rational x in R, let T_{x} denote the line segment joining p to (x,0). Let T be the union of all such T_{x}. Then T is locally connected only at p. However, it isn’t locally connected at any other point (if x is distinct from p and belongs to T, choose a neighbourhood of x (open in R^2) disjoint from p which doesn’t intersect the x-axis. The intersection of this neighbourhood with T is homeomorphic with the countable union of open intervals in R (this we leave you to check in the exercises on homeomorphisms (see next section))and therefore cannot be connected. Therefore, no neighbourhood smaller than this neighbourhood can be connected). Therefore, T cannot be weakly locally connected by theorem 5.

5. See the infinite broom (or the broom space). This is an example of a space which is weakly locally connected at a particular point, but not locally connected at that point. The reader is directed to Wolfram MathWorld for a definition of the broom space. See also the section (in this article) on ‘weakly locally connected space’.

## ExercisesEdit

**Easy Questions**

1. Prove that if X is locally path connected, then X is locally connected.

2. Prove that X has precisely one component if and only if X is connected. What is this component?

3. If f is a continuous, surjective map from a topological space X to a topological space Y, does it follow that Y is locally connected if X is locally connected? What conditions can you impose for this statement to be true, i.e. the continuous image of a locally connected space is locally connected (if your answer was yes to the first question, ignore the second; however, if your answer is yes you must prove it)?

4. a) Prove that the set I X I in the dictionary order topology is locally connected (Hint: What familiar connected subspace of R are basis elements for the topology on I X I homeomorphic to?)

b*) Find the path components of this space (Hint: First show that this space is not path connected by obtaining a contradiction and using the intermediate value theorem (assume [0,1] is connected)

c) Use your answer to b), to determine whether or not the space I X I is locally path connected.

5. Prove theorem 2

**Normal Questions**

**In these questions, we will verify whether local connectedness is preserved under products and other such operations. The reader is expected to be familiar with the product topology.**

6. a) Consider the set X = (-1,0) U (0,1) which is clearly not connected. Prove that this set is locally connected. Also find the components of this space (Hint: When proving that this space is locally connected, assume that (0,1) is connected. Then contruct a homeomorphism from (0,1) to (-1,0) to show that (-1,0) is connected)

b) Apply a theorem to show that the path components of this space are equal to the components of this space. Prove any assumptions that you make.

c) Let E be an elementary subset of R (this term is derived from measure theory where E is said to be an elementary subset of R, if it is the finite union of intervals (not necessarily open), prove that E is locally connected by assuming that each interval is connected.

7. Suppose we take the finite product of X with itself. Is this new space locally connected? (Hint: Use the fact that the product of connected spaces is always connected)

8. Now, what if we take the infinite product of X with itself given the product topology; call this new space Y. We will check whether Y is locally connected or not in question 8. We leave the reader to have a think about this before he reads question 8.

**See question 7 and think before reading question 8**:

9. a) Are basis elements for the product topology on Y connected? (Hint: The projection map is continuous; therefore the projection of a connected space is always connected)

b) Possibly a) gives us a conjecture whether Y is locally connected or not. Depending on your answer to a), prove or disprove that Y is locally connected. (Hint: If your answer to a) was yes, find at least one connected neighbourhood of a point (note that a neighborhood of a point need NOT BE a basis element). If you answer to a) was no, proceed by contradiction:

If x is a point of Y and V is a neighborhood of x, suppose U is a connected neighborhood of x contained in V. Then U contains a basis element about X. Note that the projection map is continuous so that the image of U under each projection should be connected. Obtain a contradiction)

c) Is Y weakly locally connected? Justify your answer (Hint: The idea is similar to part b))

**Research Questions**

**You will be asked to formulate hypothesis based on your knowledge**:

10. Like components of a topological space are always closed, do you think path components satisfy a similar property?

**We will analyse this question (see now question 9 and think before answering 10)**

11. a) Is it possible to determine the path components of a connected space (without knowing what the space is)? Is it possible to determine the path components of a path connected space without knowing what the space is?

b) The topologist’s sine curve is not path connected. Assuming this determine its path components and analyze the properties these path components satisfy (determine whether they are closed or open or neither).

c) From 10.b), you should be able to answer question 9. As a question which we will leave the reader to think about:

12. Are components of a topological space necessarily locally connected?

13. Are path components of a topological space necessarily locally path connected? (Hint: This question is a little trickier than 7; you might have to construct your own space (a subspace of R^2))

14. a) Prove that homeomorphisms preserve path connectedness.

b) Deduce that the topologist’s sine curve and the comb space are not homeomorphic.

**The following question is a research type question**:

15*. Like there are the components of a space, can you invent a similar equivalence relation that determines the ‘local components’ of a space. These ‘local components’ should be the largest locally connected subsets of the space.

**Example Type Questions**

16. Give an example where the path components of a space and the components of a space are not equal. Give a different example where the path components and quasicomponents of a space are not equal.

17. Find an example of a locally connected space that is not locally path connected.

18*. Give an example where the quasicomponents of a space and the components of the space are not equal.

19. If X is locally homeomorphic with Y and X is locally connected, need Y be locally connected? Justify your answer.

**Questions on weakly locally connected spaces**

20. Formulate a new notion called weak local path connectedness that uses the same idea as the definition of a weakly locally connected space.

a) Prove that every weakly locally path connected space is locally path connected

b) Give an example of a weakly locally path connected space that isn’t locally path connected (at a particular point).

21. As in theorem 2, characterize weakly locally connected spaces in terms of components (Hint: Remember theorem 5)

**Questions on Homeomorphisms**

22. In the section on ‘applications of the theorems’, check the details of 4.

23. Prove that if X has ‘n’ components, and Y is homeomorphic to X, Y has ‘n’ components. Give an example where f is a continuous surjective map from X to Y, and X has ‘n’ components but Y doesn’t.

24. a) Suppose p is a continuous, closed surjective map such that p^(-1) {y} is compact in X for each element y of Y (p: X->Y) (the reader who is not familiar with compactness is referred to Wikipedia). Prove that if Y is locally connected, so is X.

b) Prove that if Y is connected so is X

c) If X is locally connected, need Y be locally connected?

**Difficult question**

25. Is every connected Hausdorff space with more than one point uncountable (Hint: Every connected metric space with more than one point in uncountable; prove this assertion. See what difficulty arises with this proof when one considers a Hausdorff space. Remember that metric spaces are always first countable, normal, paracompact etc… whereas Hausdorff spaces aren’t)?