Path Connectedness of Open and Connected Sets in Euclidean Space

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 ballB \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, ... \}\$.