Equivalence of Definitions of Locally Connected Space

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Theorem

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

Definition 1: using Local Bases

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

Definition 2: using Neighborhood Bases

A topological space $T = \struct{S, \tau}$ is locally connected if and only if $T$ is weakly locally connected at each point of $T$.

Definition 3: using (Global) Basis

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

Definition 4: using Open Components

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

$\Box$

Definition 2 implies Definition 1

Let $T$ be weakly locally connected at each point of $T$.

That is, each point of $T$ has a neighborhood basis consisting of connected sets of $T$.

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

Let $\mathcal B_x = \set{ W \in \tau : x \in W, W \text{ is connected in } T}$.

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

Let $V = \map {\operatorname{Comp}_x} U$ denote the 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 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 component:

$W \subseteq \map {\operatorname{Comp}_y} U = \map {\operatorname{Comp}_x} U = V$

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 $V \in \mathcal B_x$.

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

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

$\Box$

Definition 1 implies Definition 3

Let each point $x$ of $T$ have a local basis $\mathcal D_x$ consisting entirely of connected sets in $T$.

By definition of local basis, each of these connected sets in $\mathcal D_x$ is open in $T$.

Consider the set $\displaystyle \mathcal D = \bigcup_{x \mathop \in S} \mathcal D_x$.

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

Since each $\mathcal D_x$ consists entirely of connected sets, $\mathcal D$ also consists entirely of connected sets.

$\Box$

Definition 3 implies Definition 1

Let $T$ have a basis $\mathcal B$ consisting of connected sets in $T$.

For each $x \in S$ we define:

$\mathcal B_x = \set {B \in \mathcal B: x \in B}$

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

As each element of $\mathcal B_x$ is also an element of $\mathcal B$, it follows that $\mathcal B_x$ is also formed of connected sets.

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

$\Box$

Definition 3 implies Definition 4

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

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

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

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

$\Box$

Definition 4 implies Definition 3

Let the components of the open sets of $T$ are also open in $T$.

Let $\mathcal B = \set {U \in \tau : U \text{ is connected in } T}$.

Let $U$ be open in $T$.

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

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

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

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

By definition, $\mathcal B$ is an basis for $T$.

$\blacksquare$


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