Contents

Contents

Definition

A topological space is called locally path-connected if it has a basis of path-connected neighbourhoods. In other words, if for every point $x$ and neighbourhood $V \ni x$, there exists a path-connected neighbourhood $U \subset V$ that contains $x$.

Properties

Lemma

Let $X$ be a locally path connected space. Then the path connected component $P_x \subset X$ over any point $x \in X$ is an open set.

Proof

It is sufficient to show that every point $y \in P_x$ has a neighbourhood $U_y$ which is still contained in $P_x$. But by local path connectedness, $y$ has a neighbourhood $V_y$ which is path connected. It follows by concatenation of paths that $V_y \subset P_x$.

A locally path-connected space is connected if and only if it is path-connected.

Proposition

The connected components of a locally path-connected space are the same as its path-connected components.

Proof

A path connected component is always connected (this lemma), and in a locally path-connected space is it also open (lemma ). This means that every path-connected component is also connected.

Conversely, it is now sufficient to see that every connected component is path-connected. Suppose it were not, then it would be covered by more than one disjoint non-empty path-connected components. But by lemma these would be all open. This would be in contradiction with the assumption that $U$ is connected. Hence we have a proof by contradiction.

Examples

Examples

(Euclidean space is locally path-connected)

For $n \in \mathbb{N}$ then Euclidean space $\mathbb{R}^n$ (with its metric topology) is locally path-connected, since each open ball is path-connected topological space.

Similarly the open intevals?, closed intervals and half-open intervals, regarded as topological subspaces of the Euclidean real line, are locally path connected.

Example

(open subspace of locally path-connected space is locally path connected)

Every open subspace of a locally path connected topological space is itself locally path connected.

Example

(circle is locally path-connected)

$S^1 = \left\{ x \in \mathbb{R}^2 \;\vert\; {\Vert x\Vert} = 1\right\} \subset \mathbb{R}^2$

is locally path connected.

Proof

By definition of the subspace topology and the defining topological base of the Euclidean plane, a base for the topology of $S^1$ is given by the images of open intervals under the local homeomorphism

$(cos(-), sin(-)) \;\colon\; \mathbb{R}^1 \to S^1 \,.$

But these open intervals are locally path connected by example , and in fact they are, evidently path-connected topological space.

The condition is a necessary assumption in the

in the form of the condition for

Last revised on March 31, 2019 at 17:39:30. See the history of this page for a list of all contributions to it.