> [!theorem] > > Let $A = (a, b), B = (c, d)$ be open intervals where $-\infty \le a < b \le \infty$ and $-\infty \le c < d \le \infty$, then if $A \cap B \ne \emptyset$, $A \cup B = (a, d)$ is another open interval. > > *Proof*. Assume without loss of generality that $a \le c$. If $A \cap B \ne \emptyset$, then $c \in (a, b)$ implies that $c < b$. Then $A \cup B = (a, d)$. > [!theorem] > > Let $\bracs{(a_i, b_i)}_1^n$ be a collection of open intervals and $U = \bigcup_{i = 1}^{n}(a_i, b_i)$. Then we can rewrite > $ > U = \bigcup_{i = 1}^{m}(c_i, d_i) > $ > as a disjoint union. > > *Proof*. Suppose that $\bracs{(a_i, b_i)}_1^n$ is disjoint, then $U$ is already a disjoint union. > > If $n \ge 2$ and there exists $j, k \in [1, n], j \ne k$ such that $(a_j, b_j)$ and $(a_k, b_k)$ are not disjoint. Then replace them with $(a_j, b_j) \cup (a_k, b_k)$ as a single open interval. This reduces the number of open intervals in the sequence by $1$. > > This process will terminate until $n = 1$ or when all intervals are disjoint, in both cases $U$ has been rewritten as a disjoint union of open intervals. > [!theorem] > > Let $U \in \topo_\real$ be an [[Open Set|open set]], then > $ > U = \bigcup_{n \in \nat}(a_n, b_n) \quad -\infty \le a_n < b_n \le \infty > $ > is a countable disjoint union of open intervals. > > *Proof*. Let $x \sim y$ if there exists an open interval $(a, b)$ contained in $U$ such that $x, y \in (a, b) \subset U$. This relation is reflexive as for any $x \in U$, > $ > \exists \varepsilon> 0: x \in B(x, \varepsilon_x) \subset U > $ > and symmetric. Suppose that $x \sim y$ and $y \sim z$ where $x, y \in (a, b)$ and $y, z \in (c, d)$, then since $y \in (a, b) \cap (c, d)$, their intersection is non-empty and $(a, b) \cup (c, d)$ is an interval in $U$ containing $x, z$. Since $\sim$ is reflexive, symmetric, and transitive, $\sim$ is an [[Equivalence Relation|equivalence relation]]. > > Let $[x]$ be the [[Equivalence Class|equivalence class]] corresponding to $x$, then for any $y, z \in [x]$, > $ > \begin{align*} > \exists (a, b): y, z \in (a, b) &\Rightarrow [y, z] \subset (a, b) \\ > &\Rightarrow y, z \sim s \forall s \in [y, z] \\ > &\Rightarrow [y, z] \subset [x] > \end{align*} > $ > and $[x]$ is an interval. Moreover, for any $y \in [x]$, $y \sim y$ implies that $\exists (a, b): y \in (a, b) \subset [x]$ and $[x]$ is open. Therefore $[x]$ is an open interval. > > Since $[x]$ is open, for any $y \in [x]$, > $ > \exists \varepsilon > 0: (x - \varepsilon, x + \varepsilon) \subset [x] > $ > and since the [[Rational Numbers|rational numbers]] are [[Dense|dense]], $\exists q \in (x, x + \varepsilon)$ such that $x, q \in (x - \varepsilon, x + \varepsilon)$ and $x \sim q$. > > Let $S$ be the complete set of representatives, then since each element of $S$ is equivalent to some rational number, $S$ is countable. Moreover, since equivalence classes are either disjoint or equal, this partitions > $ > U = \bigcup_{x \in S}[x] > $ > into a countable union of open intervals.