Totally disconnected spaces have very small connected pieces. Local connectedness asks for the opposite behaviour near each point: small connected neighbourhoods should be available inside every open neighbourhood. Local forms of connectedness are not the same as global connectedness. A space may be locally connected without being connected, and a connected space may fail to be locally connected. The focus keyword local forms of connectedness appears here because the lesson separates local structure from the global one-piece property. :::definition[Locally Connected Metric Space] Let $(X,d)$ be a metric space. Then $X$ is said to be $locally connected$ if for each $x\in X$, there exists a base of connected neighbourhoods of $x$. Equivalently, the family of all open connected subsets of $X$ is a base for the open subsets of $X$. Thus, whenever $O$ is open in $X$ and $x\in O$, there exists an open connected set $U$ such that $x\in U\subseteq O$. ::: :::example[Locally Connected but Not Connected] Let $X=\{x\in\mathbb{R}:a<x<b\ \text{or}\ c<x<d\}$, where $b<c$. Then $X=(a,b)\cup(c,d)$. Each point of $X$ has a small connected open interval around it, so $X$ is locally connected. However, $X$ is the union of two nonempty disjoint open subsets. Hence $X$ is not connected. $\square$ ::: :::example[The Discrete Two-Point Space] Let $X_0=\{0,1\}$ with the discrete metric. The singleton subsets $\{0\}$ and $\{1\}$ are open and connected, and they form a base for the topology of $X_0$. Therefore $X_0$ is locally connected. But $X_0=\{0\}\cup\{1\}$ is a separation, so $X_0$ is not connected. $\square$ ::: :::example[Connected but Not Locally Connected] Let $Y=\{(x,0):1/2<x\leq 1\}\cup\{(x,y):y=x/n,\ 0\leq x\leq 1,\ n\in\mathbb{N}\}$. The set $Y$ is connected because it lies between a connected union of line segments and its closure. However, $Y$ is not locally connected. Near a point of the horizontal segment, such as $(3/4,0)$, small relative neighbourhoods meet infinitely many slanting line segments and do not contain connected open neighbourhoods inside $Y$. $\square$ ::: :::theorem[Local Connectedness and Components of Open Sets] Let $(X,d)$ be a metric space. Then $X$ is locally connected if and only if the connected components of open subsets of $X$ are open in $X$. ::: :::proof Given that $(X,d)$ is a metric space. To prove that $X$ is locally connected if and only if the components of open sets are open. First assume that $X$ is locally connected. Let $G$ be open in $X$, and let $C$ be a connected component of $G$. Let $x\in C$. Since $G$ is open and $x\in G$, local connectedness gives an open connected set $U$ such that $x\in U\subseteq G$. Since $U$ is connected and contains $x$, it is contained in the connected component $C$ of $G$. Thus $x\in U\subseteq C$. Therefore every point of $C$ has an open neighbourhood contained in $C$. Hence $C$ is open in $X$. Conversely, assume that connected components of open subsets of $X$ are open in $X$. Let $O$ be open in $X$ and let $x\in O$. Let $C$ be the connected component of $x$ in $O$. By assumption, $C$ is open in $X$. Also, $C$ is connected and $x\in C\subseteq O$. Therefore every open neighbourhood of $x$ contains an open connected neighbourhood of $x$. Hence $X$ is locally connected. $\square$ ::: :::theorem[Open Subsets of the Real Line] Let $O$ be a nonempty open subset of $\mathbb{R}$. Then $O$ is a union of an at most countable family of pairwise disjoint open intervals. ::: :::proof Given that $O$ is a nonempty open subset of $\mathbb{R}$. To prove that $O$ is a union of an at most countable family of pairwise disjoint open intervals. Since $\mathbb{R}$ is locally connected, the connected components of $O$ are open in $\mathbb{R}$. Each connected component of $O$ is a connected subset of $\mathbb{R}$, so each component is an interval. Since each component is open, each component is an open interval. The connected components of $O$ are pairwise disjoint and their union is $O$. Each nonempty open interval contains a rational number. Since the rational numbers are countable and the components are pairwise disjoint, there can be at most countably many components. Hence $O$ is a union of an at most countable family of pairwise disjoint open intervals. $\square$ ::: :::exercise [1] Define local connectedness. [2] Give an example of a locally connected space that is not connected. [3] State the component criterion for local connectedness. [4] Explain why every open subset of $\mathbb{R}$ is a countable union of disjoint open intervals. ::: :::answer [1] Every point has a base of connected neighbourhoods. [2] The union $(a,b)\cup(c,d)$ with $b<c$ is locally connected but disconnected. [3] A metric space is locally connected if and only if components of open subsets are open. [4] Components of open subsets of $\mathbb{R}$ are open intervals, and each contains a rational number. ::: :::faq[Frequently Asked Questions] Q: Does locally connected imply connected? A: No. It is a local condition near each point. Q: Does connected imply locally connected? A: No. The comb-type example is connected but not locally connected. Q: Why are components of open sets important? A: They reveal whether connected neighbourhoods are available locally. ::: :::call-to-action[Continue to Paths] subtitle: Local forms of connectedness prepare the transition from connected sets to path-connected sets. button: Next Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/paths-and-path-connected-sets button-ghost: Previous Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/totally-separated-metric-spaces :::