Linear Connectedness Implies Local Connectedness

In this note we discuss two notions of connectedness in the context of a metric space. These are Linear Connectedness and Local Connectedness. We wish to prove that linear connectedness implies local connectedness (however the converse is not true).

Linear Connectedness or bounded turning is a geometric condition that is defined as follows:

Suppose $(X, d)$ is a metric space. We say it is $L$-linearly connected for some $L \geq 1$ if for all $x, y \in X$ there exists a compact, connected set $J \ni x, y$ of diameter less than or equal to $Ld(x, y)$.

In order to understand local connectedness, let us first define the notion of a neighbourhood.

If $X$ is a topological space and $p$ is a point in $X$, then a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ containing $p$,

$$
p \in U \subseteq V \subseteq X
$$

Local connectedness in a metric space is defined as follows.

Suppose $(X, d)$ is a metric space. Let $x \in X$ and $V \ni x$ a neighbourhood of $x$. Then there exists a connected neighbourhood $U$ such that $x \in U \subseteq V$.

We wish to show that if a metric space $X$ is linearly connected then it is locally connected. In particular, suppose $x \in X$ is any point and $V$ is any neighbourhood of $x$ in $X$. Then there exists a connected neighbourhood $U$ of $x$ that is contained in $V$.

If $V$ is a neighbourhood for $x$, then $x \in \operatorname{int}(V)$ and then there exist $\epsilon>0$ such that $B(x, \epsilon) \subseteq V$.

Recall that, by assumption, $(X, d)$ is $L$-linearly connected. Choose a point $y \in B(x, \epsilon)$ such that $d(x, y) < \frac{\epsilon}{L}$. If there is no such point then $B(x,\frac{\epsilon}{L}) = \{x\}$ is a connected open set containing $x$ and we are done.

Otherwise proceed as follows.

Since $(X, d)$ is $L$-linearly connected, there exists a compact, connected set $J \ni x, y$ of diameter less than or equal to $L \times d(x, y)$. Hence $diam(J) < L \times \frac{\epsilon}{L} = \epsilon$. This implies $ J \subseteq B(x, \epsilon)$. Therefore we have a connected compact set $J$ that contains $x$ and is contained in $V$.

We essentially proved that every neighbourhood of $x \in X$ contains a connected set $J$ that also contains $x$. This property is known as weakly locally connected or connected im-kleinan.

The precise definition goes as follows:

$X$ is connected im kleinen at $x$ iff each open neighbourhood $U$ of $x$ contains an open neighbourhood $V$ of $x$ such that any pair of points in $V$ lie in some connected subset of $U$.

or equivalently $X$ is connected im kleinen at $x$ iff each open neighbourhood $U$ of $x$ contains a connected neighbourhood $V$ of $x$ (not necessarily open).

It is a theorem that:

If $X$ is connected im kleinen at each point, then $X$ is locally connected.

Hence we are done!

If $X$ is connected im kleinen at all points $x \in X$ then $X$ is locally connected.

Suppose $x \in X$ be any point. Let $U$ be any neighbourhood of $x$ in $X$. We wish to show that there is a open set $V$ contained in $U$ that contains $x$.