> [!theorem] > > $ > |b - a| \equiv d(a, b) = \sqrt{\sum_{k=1}^{n}(a_k - b_k)^2} > $ > The [[Real Coordinate Space|real space]] $\real^n$ with [[Euclidean Space|Euclidean distance]] as its distance function is a [[Metric Space|metric space]]. > [!definition] > > A [[Set|set]] $U \subseteq \real^n$ is an [[Open Set|open set]] if > $ > \forall a \in U, \exists \varepsilon > 0: \bracs{x: |x - a|< \varepsilon} \subseteq U > $ > [!definition] > > Let $\topo$ be the set of all open sets. $(\real^n, \topo)$ is a [[Topological Space|topological space]] and is called the **standard topology on $\real^n$**. > > *Proof*. Let $A, B \in \topo$. Then $A \cup B$ is obviously another open set, and therefore $A \cup B \in \topo$. Let $x \in A \cap B$. Then > $ > \begin{align*} > \exists \varepsilon_a > 0: \bracs{x: |x - a| < \varepsilon_a} \subseteq A \\ > \exists \varepsilon_b > 0: \bracs{x: |x - a| < \varepsilon_b} \subseteq B > \end{align*} > $ > Let $\varepsilon = \min(\varepsilon_a, \varepsilon_b)$, and assume without loss of generality that $\varepsilon_a \le \varepsilon_b$. Then, > $ > \bracs{x: |x - a| < \varepsilon_a} \subseteq \bracs{x: |x - a| < \varepsilon_b} \subseteq B > $ > and since $\bracs{x: |x - a| < \varepsilon_a} \subseteq A$, $\bracs{x: |x - a| < \varepsilon_a} \subseteq A \cap B$. Therefore $A \cap B$ is an open set. > > $\real^n$ is an open set, and $\emptyset$ is an open set as no elements are in it. > [!theorem] > > The standard topology on $\real^n$ is a [[Hausdorff Space|Hausdorff space]]. > > *Proof*. Let $a, b \in \real^n$, $a \ne b$. Then $|a - b| = \varepsilon$. Let > $ > A = \bracs{x: |x - a|< \varepsilon/3} \quad B = \bracs{x: |x - b| < \varepsilon/3} > $ > Let $x \in A \cap B$, then > $ > |a - b| = |a - x + x - b| \le |x - a| + |x - b| = \frac{2\varepsilon}{3} > $ > Contradiction. Therefore $A \cap B = \emptyset$, and $\real^n$ is a Hausdorff space. > [!theorem] > > Every [[Open Set|open set]] in $\real$ is a **countable** **disjoint** union of open intervals.