Locally connected space Totally Explained

Everything about Locally Connected Space totally explained

In topology and other branches of mathematics, 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's 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).

Examples

1. If a space has a basis consisting entirely of connected sets, then it's locally connected (to see this, let x be a point of the space. Then for each neighbourhood V of x, there's 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's 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 can't 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 ť 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 isn't locally connected. The fact that the topologist's sine curve is connected follows from: ť a) The set S = f((0,1]) is connected since it's the image of a connected space under a continuous map.

   b) The closure of a connected space is connected. The space isn't locally connected at any point in the set B = [Closure(S)] – S. It is locally connected at every other point though.

Local path connectedness

A topological space, X, is locally path connected, if for each point x, and each neighborhood V of x, there's 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 isn't locally path connected is the comb space: if K = × 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.

Theorems

In this section, we'll 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 iff 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're disjoint, nonempty (P is nonempty since x is in P; Q is nonempty, because we're 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's 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's a path from z to y. Since there's 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's 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's a path from z to x, then there would be a path from x to y since there's already a path from z to y. This can't be since y is in U−A. Therefore, there's 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.

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 R² is path connected.
2. The set of rational numbers Q isn't 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.

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