Locally connected implies disjoint union of open connected sets

Question

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.

Answer 1

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.

Answer 2

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.