Proof of Euclidean Open Box Criterion in $\mathbb{R}^n$

Assume first that $U$ is open in the Euclidean sense, and let $x=(x_1,\dots,x_n)\in U$. By Open Subset of Euclidean Space, there exists $r>0$ such that every point $y=(y_1,\dots,y_n)\in\mathbb{R}^n$ satisfying

belongs to $U$.

Set

Let $y=(y_1,\dots,y_n)\in\mathbb{R}^n$ satisfy

for every $i\in\{1,\dots,n\}$. Then

so

for every $i$. Therefore

Hence $y\in U$. This proves the coordinate-box condition.

Conversely, assume that for every $x=(x_1,\dots,x_n)\in U$ there exists $\delta>0$ such that every $y=(y_1,\dots,y_n)\in\mathbb{R}^n$ satisfying

for every $i\in\{1,\dots,n\}$ belongs to $U$. Let $x\in U$, and choose such a $\delta>0$. If $y\in\mathbb{R}^n$ satisfies

then in particular

for each $i$, so

and therefore

for every $i$. By hypothesis, this implies $y\in U$.

Thus for every $x\in U$ there exists a Euclidean neighborhood of $x$ contained in $U$, so $U$ is open in the Euclidean sense. The two conditions are equivalent.