Equivalence of Definitions of Locally Path-Connected Space

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Theorem

The following definitions of the concept of Locally Path-Connected Space are equivalent:


Definition 1: using Local Bases

A topological space $T = \struct{S, \tau}$ is a locally path-connected space if and only if each point of $T$ has a local basis consisting of path-connected sets in $T$.

Definition 2: using Neighborhood Bases

A topological space $T = \struct{S, \tau}$ is a locally path-connected space if and only if each point of $T$ has a neighborhood basis consisting of path-connected sets in $T$.

Definition 3: using (Global) Basis

A topological space $T = \struct {S, \tau}$ is a locally path-connected space if and only if it has a basis consisting of path-connected sets in $T$.

Definition 4: using Open Path Components

A topological space $T = \struct{S, \tau}$ is a locally path-connected space if and only if the path components of open sets of $T$ are also open in $T$.


Proof

Definition 1 implies Definition 2

Let each point of $T$ have a local basis consisting entirely of path-connected sets in $T$.

From Local Basis for Open Sets Implies Neighborhood Basis of Open Sets, it follows directly that:

each point of $T$ has a neighborhood basis consisting entirely of path-connected sets in $T$.

$\Box$


Definition 2 implies Definition 1

Let each point of $T$ have a neighborhood basis consisting of path-connected sets in $T$.

Let $x \in S$ and $x \in U \in \tau$.

Let $\BB_x = \set {W \in \tau : x \in W, W \text { is path-connected in } T}$.

By definition of local basis, we have to show that there exists a path-connected open set $V \in \tau$ with $x \in V \subset U$.

Let $V = \map {\operatorname {PC}_x} U$ denote the path component of $x$ in $U$.

From Open Set in Open Subspace, it suffices to show that $V$ is open in $U$.

By Set is Open iff Neighborhood of all its Points, we may do this by showing that $V$ is a neighborhood in $U$ of all of its points.


Let $y \in V$.

By assumption, there exists a path-connected neighborhood $W$ of $y$ in $T$ with $W \subset U$.

By Neighborhood in Topological Subspace, $W$ is a neighborhood of $y$ in $U$.

By definition of path component:

$W \subset \map {\operatorname {PC}_y} U = \map {\operatorname {PC}_x} U = V$

By Neighborhood iff Contains Neighborhood, $V$ is a neighborhood of $y$ in $U$.

Because $y$ was arbitrary, $V$ is open in $U$.

From Open Set in Open Subspace, $V$ is open in $T$.

Hence $V \in \BB_x$.

Because $U$ was arbitrary, $\BB_x$ is a local basis of $x$ consisting of (open) path-connected sets.

Since $x$ was arbitrary, then each point of $T$ has a local basis consisting entirely of path-connected sets in $T$.

$\Box$


Definition 1 implies Definition 3

Let each point $x$ of $T$ have a local basis $\DD_x$ consisting entirely of path-connected sets in $T$.

By definition of local basis, each of these path-connected sets in $\DD_x$ is open in $T$.

Consider the set $\ds \DD = \bigcup_{x \mathop \in S} \DD_x$.

From Union of Local Bases is Basis, $\DD$ is a basis for the topology $\tau$.

Since each $\DD_x$ consists entirely of path-connected sets, $\DD$ also consists entirely of path-connected sets.

$\Box$


Definition 3 implies Definition 1

Let $T$ have a basis $\BB$ consisting of path-connected sets in $T$.

For each $x \in S$ we define:

$\BB_x = \set {B \in \BB: x \in B}$

From Basis induces Local Basis, $\BB_x$ is a local basis.

As each element of $\BB_x$ is also an element of $\BB$, it follows that $\BB_x$ is also formed of path-connected sets.

Thus, for each point $x \in S$, there is a local basis which consists entirely of path-connected sets.

$\Box$


Definition 3 implies Definition 4

Let $T$ have a basis consisting of path-connected sets in $T$.

Let $U$ be an open subset of $T$.

From Open Set is Union of Elements of Basis, $U$ is a union of open path-connected sets in $T$.

From Open Set in Open Subspace and Path-Connected Set in Subspace, $U$ is a union of open path-connected sets in $U$.

From Path Components are Open iff Union of Open Path-Connected Sets, the path components of $U$ are open in $U$.

From Open Set in Open Subspace then the path components of $U$ are open in $T$.

$\Box$


Definition 4 implies Definition 3

Let the path components of open sets of $T$ be also open in $T$.

Let $\BB = \set {U \in \tau : U \text{ is path-connected in } T}$.

Let $U$ be open in $T$.

By assumption, the path components of $U$ are open in $T$.

From Path-Connected Set in Subspace, the path components of $U$ are path-connected in $T$.

By the definition of the path components of a topological space, $U$ is the union of its path components.

Hence $U$ is the union of open path-connected sets in $T$.

By definition, $\BB$ is an basis for $T$.

$\blacksquare$


Also see


Sources

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