Proof of Products of Euclidean Open Sets are Open

Throughout, $\mathbb{R}$ is the real numbers, which together with its order is in particular an ordered field, so the claims of Elementary Order Arithmetic in an Ordered Field apply to it; openness always means openness in the Euclidean sense.

Claim 1. Let $x=(x_1,\dots,x_p)\in\mathbb{R}^p$ and take $r=1$, which satisfies $r>0$ by claim 6 of Elementary Order Arithmetic in an Ordered Field. Every point $y\in\mathbb{R}^p$ with $\sum_{i=1}^p(y_i-x_i)^2<r^2$ belongs to $\mathbb{R}^p$, since it was taken in $\mathbb{R}^p$. So the defining condition of openness holds at every point of $\mathbb{R}^p$, and $\mathbb{R}^p$ is open in $\mathbb{R}^p$.

Claim 2. Let $U\subseteq\mathbb{R}^p$ and $V\subseteq\mathbb{R}^q$ be open, and let $x\in U\times V$. By claim 1 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space the concatenation map is a bijection whose inverse reads off coordinates, so $x$ corresponds to exactly one pair $(\xi,\eta)$, and $x\in U\times V$ means $\xi\in U$ and $\eta\in V$, where $\xi_i=x_i$ for $1\le i\le p$ and $\eta_j=x_{p+j}$ for $1\le j\le q$.

Since $U$ is open, the forward implication of Euclidean Open Box Criterion in $\mathbb{R}^n$, applied in $\mathbb{R}^p$, provides a real $\delta_1>0$ such that every $\xi'\in\mathbb{R}^p$ satisfying $\xi_i-\delta_1<\xi'_i<\xi_i+\delta_1$ for all $i\in\{1,\dots,p\}$ belongs to $U$. Since $V$ is open, the same criterion applied in $\mathbb{R}^q$ provides a real $\delta_2>0$ such that every $\eta'\in\mathbb{R}^q$ satisfying $\eta_j-\delta_2<\eta'_j<\eta_j+\delta_2$ for all $j\in\{1,\dots,q\}$ belongs to $V$. By claim 9 of Elementary Order Arithmetic in an Ordered Field there is a real $\delta$ with $\delta\le\delta_1$, $\delta\le\delta_2$, and $\delta$ equal to $\delta_1$ or to $\delta_2$; in either case $\delta>0$.

We first record, for any real $t$ and any real $\delta'$ with $\delta\le\delta'$, that

$$t-\delta'\le t-\delta\qquad\text{and}\qquad t+\delta\le t+\delta'.$$

Indeed, either $\delta=\delta'$, and both relations are equalities; or $\delta<\delta'$, in which case claim 4 of Elementary Order Arithmetic in an Ordered Field gives $-\delta'<-\delta$ and claim 1 of that lemma, adding $t$, gives $t-\delta'<t-\delta$ and $t+\delta<t+\delta'$.

Now let $y\in\mathbb{R}^{p+q}$ satisfy $x_k-\delta<y_k<x_k+\delta$ for every $k\in\{1,\dots,p+q\}$, and let $(\xi',\eta')$ be the pair corresponding to $y$, so that $\xi'_i=y_i$ for $1\le i\le p$ and $\eta'_j=y_{p+j}$ for $1\le j\le q$. Fix $i\in\{1,\dots,p\}$. Applying the recorded relations with $t=\xi_i$ and $\delta'=\delta_1$, and using $x_i=\xi_i$ and $y_i=\xi'_i$, we get

$$\xi_i-\delta_1\le\xi_i-\delta<\xi'_i<\xi_i+\delta\le\xi_i+\delta_1,$$

so $\xi_i-\delta_1<\xi'_i<\xi_i+\delta_1$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. As $i$ was arbitrary, $\xi'\in U$. The same argument with $t=\eta_j$, $\delta'=\delta_2$ and the coordinates $x_{p+j}=\eta_j$, $y_{p+j}=\eta'_j$ gives $\eta_j-\delta_2<\eta'_j<\eta_j+\delta_2$ for every $j\in\{1,\dots,q\}$, so $\eta'\in V$. Hence $y$, the point corresponding to $(\xi',\eta')$, lies in $U\times V$.

Thus every $x\in U\times V$ admits a real $\delta>0$ with the stated box property, and the reverse implication of Euclidean Open Box Criterion in $\mathbb{R}^n$, applied in $\mathbb{R}^{p+q}$, shows that $U\times V$ is open in $\mathbb{R}^{p+q}$.