Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous

Let $n$ and $m$ be natural numbers and let $\mathbb{R}$ be the real numbers. Regard Euclidean space $\mathbb{R}^{n}$ as a metric space through the Euclidean distance $d_{E}$, which is a metric by $\mathbb{R}^n$ with the Euclidean Distance Is a Complete Metric Space with Coordinatewise Convergence, and regard $\mathbb{R}$ as a metric space through the metric $d_{\mathbb{R}}$ of The Absolute Value Metric on the Real Numbers.

Then the following hold.

1. (The whole space is open) $\mathbb{R}^{n}$ is an open subset of $\mathbb{R}^{n}$.

2. (Class $C^{k}$ implies class $C^{1}$) Let $U\subseteq\mathbb{R}^{n}$ be open, let $F=(F_{1},\dots,F_{m}):U\to\mathbb{R}^{m}$, and let $k$ be a natural number. If $F$ is of class $C^{k}$ on $U$, then $F$ is of class $C^{1}$ on $U$.

3. (Continuity) Let $U$, $F$ and $k$ be as in claim 2, with $F$ of class $C^{k}$ on $U$. Then for every $j$ with $1\le j\le m$ the coordinate function $F_{j}:U\to\mathbb{R}$ is continuous in the Euclidean sense at every point of $U$, and is also continuous relative to $U$ at every point of $U$, as a map from $U$ into $(\mathbb{R},d_{\mathbb{R}})$. The same conclusions hold if $F$ is instead assumed smooth on $U$.