locally path-connected space




topology (point-set topology, point-free topology)



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



Let XX be a locally path connected space. Then the path connected component P xXP_x \subset X over any point xXx \in X is an open set.


It is sufficient to show that every point yP xy \in P_x has a neighbourhood U yU_y which is still contained in P xP_x. But by local path connectedness, yy has a neighbourhood V yV_y which is path connected. It follows by concatenation of paths that V yP xV_y \subset P_x.

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


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


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 UU is connected. Hence we have a proof by contradiction.



(Euclidean space is locally path-connected)

For nn \in \mathbb{N} then Euclidean space n\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.


(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.


(circle is locally path-connected)

The Euclidean circle

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

is locally path connected.


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

(cos(),sin()): 1S 1. (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.