Path Connectedness of Open and Connected Sets in Euclidean Space

Theorem 1: If $A$ is an open and connected subset of $\mathbb{R}^n$ (with the usual topology) then $A$ is path connected.

• Proof: Let $\mathbb{R}^n$ have the usual topology and let $A \subseteq \mathbb{R}^n$ be an open and connected subset of $\mathbb{R}^n$.

• Let $x, y \in A$ and let $\mathcal R$ denote the collection of all open balls contained in $A$:

(1)

\begin{align} \quad \mathcal R = \{ B = B(x, r) : x \in A, r > 0, B(x, r) \subseteq A \} \end{align}

• Since $A$ is open and since $x \in A$ there exists an open ball contained in $\mathcal R$ that contains $x$, call it $B_1 = B_1(x, r_x)$.

• Set $B_2$ to be the union of all open balls $B \in \mathcal R$ such that $B \cap U_1 \neq \emptyset$:

(2)

\begin{align} \quad B_2 = \left \{ \bigcup_{B \in \mathcal R} B : B \cap U_1 \neq \emptyset \right \} \end{align}

• In general, for all $i \in \mathbb{N}$ $i > 1$, set $B_i$ to be the union of all open balls $B \in \mathcal R$ such that [[$ B \cap U_{i-1} \neq \emptyset:

(3)

\begin{align} \quad B_i = \left \{ \bigcup_{B \in \mathcal R} B : B \cap U_{i-1} \neq \emptyset \right \} \end{align}

• We claim that every ball $B \in \mathcal R$ is contained in some $B_i$. Suppose not. Let $\displaystyle{V = \bigcup_{i=1}^{\infty} B_i}$ and let $\displaystyle{W = \bigcup_{B \in \mathcal R^*} B}$ where $\displaystyle{\mathcal R^* = \{ B \in \mathcal R : B \not \subseteq U_i \: \forall i \in I}$. Then $V$ and $W$ are both open sets since they are the union of open balls. Moreover, $V, W \neq \emptyset$, $V \cap W = \emptyset$, and $A = V \cup W$. So, $\{ V, W \}$ is a separation of $A$ which contradictions $A$ being connected.

• So for every $B \in \mathcal R$ there exists a $i \in \{1, 2, ... \}$ such that $B \subseteq B_i$. So there exists an $n \in \{1, 2, ... \}$ such that $y \in B_n$. Moreover, $\displaystyle{A = V = \bigcup_{i=1}^{\infty} B_i}$ is such that $B_i \cap B_{i+1} \neq \emptyset$ for all $i \in \{1, 2, 3, ... \}$.

• So, from the theorem presented on the Path Connectivity of Countable Unions of Connected Sets page we have that $V = A$ is path connected. $\blacksquare$