Locally Connected and Locally Path Connected Topological Spaces

# Locally Connected and Locally Path Connected Topological Spaces

Recall from the Connected and Disconnected Topological Spaces page that a topological space $X$ is said to be connected if it is not disconnected.

Also recall from the Path Connected Topological Spaces page that a topological space $X$ is said to be path connected if for every pair of distinct points $x, y \in X$ there exists a continuous function $\alpha : [0, 1] \to X$, called a path from $x$ to $y$, such that $\alpha(0) = x$ and $\alpha(1) = y$.

Sometimes a topological space may not be connected or path connected, but may be connected or path connected in a small open neighbourhood of each point in the space. We define these new types of connectedness and path connectedness below.

 Definition: Let $X$ be a topological space and let $x \in X$. We say that $X$ is Locally Connected at $x$ if for every neighbourhood $U$ of $x$ there exists a connected neighbourhood $V$ of $x$ such that $x \in V \subseteq U$. $X$ is said to be Locally Connected on all of $X$ if $X$ is locally connected at every $x \in X$.
 Definition: Let $X$ be a topological space and let $x \in X$. We say that $X$ is Locally Path Connected at $x$ if for every neighbourhood $U$ of $x$ there exists a path connected neighbourhood $V$ of $x$ such that $x \in V \subseteq U$. $X$ is said to be Locally Path Connected on all of $X$ if $X$ is locally path connected at every $x \in X$.

For example, consider the topological space $\mathbb{R}$ with the usual topology. Consider the following topological subspace of $\mathbb{R}$ which we denote by $N$:

(1)
\begin{align} \quad N = \bigcup_{n \in \mathbb{Z}} (n, n+1) = ... \cup (-2, -1) \cup (-1, 0) \cup (0, 1) \cup (1, 2) \cup ... \end{align}

Clearly $N$ is disconnected. To see this, let $\displaystyle{A = \bigcup_{n \in \mathbb{Z}, n < 0} (n, n+1)}$ and let $\displaystyle{B = \bigcup_{n \in \mathbb{Z}, n \geq 0} (n, n+1)}$. Then it's not hard to verify that $\{ A, B \}$ is a separation of $N$.

However, $N$ is locally connected on all of $N$. To see this, let $x \in N$. Then $x \in (n, n+1)$ for some $n \in \mathbb{N}$. Let $\delta = \min \{ n - x, n + 1 - x \}$. Then $(x - \delta, x + \delta) \subseteq (n, n+1)$ is a connected neighbourhood of $x$.

Since $x$ was arbitrary, we see that $N$ is locally collected on all of $N$.

For another example, consider the following topological subspace of $\mathbb{R}$ which we denote by $X$:

(2)
\begin{align} \quad X = \bigcup_{n \in \mathbb{N}} \left \{ \left ( \frac{1}{n}, y \right ) : y \in \mathbb{R} \right \} \cup \{ (0, y) : y \in \mathbb{R} \} \cup \{ (x, 0) : x \in \mathbb{R} \} \end{align}

The following diagram illustrates this space: It's not hard to see that $X$ is path connected. However, $X$ is not locally path connected everywhere. Take a point $\mathbf{a} = (a, 0) \in X$ where $a \neq 0$: Then there will always exist an open neighbourhood of $\mathbf{a}$ that is not locally path connected as illustrated above. For an explicit open neighbourhood, $X \cap B(\mathbf{a}, \mid a \mid)$ will always be not locally path connected.