# Locally connected implies disjoint union of open connected sets

I'm struggling with the following question (from an introductory analysis course):

A metric space $E$ is said to be locally connected if for all $x \in E$, there exists $r > 0$ such that $B_r(x)$ is connected (this is the open ball with radius $r$ centered at $x$). Show that if $E$ is locally connected, then $E$ is the disjoint union of open connected sets.

(For reference, we have learned that a space $E$ is connected if $E$ and $\emptyset$ are the only sets that are both open and closed in $E$, and that $E$ is not connected $\iff$ there exists nonempty, disjoint open sets $A$ and $B$ such that $A \bigcup B = E$ ).

I get the feeling that I need to show that $E$ is a union of certain (disjoint) connected balls in $E$, but (if this were the case) I haven't been able to set up a way to determine which balls/sets I need to form the union.

I appreciate any and all help; thanks in advance.

• It doesn't affect the truth of this assertion, but the second paragraph does not give the usual definition of local connectedness. According to the definition here, every bounded and connected metric space would be locally connected. Under the usual definition, things like the topologist's sine curve are not locally connected. – Daniel Fischer Oct 30 '17 at 19:14

To say that $E$ is the disjoint union of open connected sets is not the same as to say that $E$ is the disjoint union of open balls.

The idea is to first formulate a general definition and prove a general fact about any topological space. Define a component of $E$ to be a maximal connected subset $C \subset E$, meaning that $C$ is connected but any subset properly containing $C$ is not connected. There are two things you have to prove about the components of $E$:

1. Every connected subset of $E$ is contained in some component of $E$.
2. Two distinct components of $E$ are disjoint.

Once you've proved those two things, it follows immediately that $E$ is the disjoint union of its components.

Now you bring in local connectivity. For each component $C$ of $E$ and each $x \in C$, there exists $r>0$ such that $B_r(x)$ is connected. The set $B_r(x)$ is connected and is therefore contained in some component of $E$, but $B_r(x)$ contains a point of $C$, namely $x$, and so $B_r(x)$ must be contained in $C$. This proves $C$ is open.

Thus $E$ has been decomposed into a disjoint union of connected open sets, namely its components.

Let $x\in E'$, where $E'$ is a connected component of $E$. We know $x$ is contained in a connected open subset $B_r(x)\subseteq E$. But $E'$ is a maximal connected set containing $x$ (since it is a connected component of $E$), so $x\in B_r(x)\subseteq E'$. Therefore $E'$ is open in $E$, since each $x\in E'$ has an open neighbourhood $B_r(x)\subseteq E'$. This shows that $E$ is a disjoint union of its connected components.