Proof of Euclidean Space Is Open in Itself

Throughout, the arithmetic and order of $\mathbb{R}$ are used through The Structure of ℝ as a Field and an Ordered Ring (i), The Structure of ℝ as a Field and an Ordered Ring (ii), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (ix), Rules of Arithmetic and Order in an Ordered Field (iii) and The Quotient 1/x, Order Rules for Sums, and Nesting of Closed Intervals in ℝ (i), norms and distances in $\mathbb{R}^{n}$ through Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (ii), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (v), Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square (vi) and Standard Basis Vectors of ℝ^q: Coordinates, Orthonormality and Unit Length (iii), and $(\mathbb{R}^{n},d_{E})$ and $(\mathbb{R},d_{\mathbb{R}})$ are metric spaces by $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence (i) and The Absolute Value Metric on the Real Numbers (i).

By The Open Subsets of a Metric Space Form a Topology (i), the open subsets of the metric space $(\mathbb{R}^{n},d_{E})$ form a topology on $\mathbb{R}^{n}$, and a topology on a set contains the whole set by Topological Space (ii). Hence $\mathbb{R}^{n}$ is open in $(\mathbb{R}^{n},d_{E})$, which is (i).