I came across this while studying some topology - my school uses the definition of a set $U \subset \mathbb{R}^n$ being open as there being an open rectangle $R$ for all $x \in U$ such that $x \in R$ and $R \subset U$.
However, from reading any textbook, I see that the definition is normally the same as above, except we look for open balls, so I'm trying to prove that these definitions are equivalent. I managed to construct open balls that contain the point in question given an open rectangle, however, I have trouble constructing an open rectangle that is a subset of a given open ball that contains the point in question. Any help on how to succinctly do so?
This follows from a standard inequality: for $x=(x_1,...,x_n)\in\mathbb{R^n}$ we have $||x||\leq\sum_{i=1}^n |x_i|$ where $||x||$ is the Euclidean norm.
So now suppose $x=(x_1,...,x_n)$ is a point in an open ball $B$. Without loss of generality $x$ is the center of $B$, otherwise just build a smaller ball with center in $x$ which is contained in $B$. Let's say the radius of $B$ is $r$. Then $B=\{y: ||x-y||<r\}$. Now I say the rectangle $(x_1-\frac{r}{n},x_1+\frac{r}{n})\times...\times (x_n-\frac{r}{n},x_n+\frac{r}{n})$ is contained in $B$. Indeed, if $y=(y_1,...,y_n)$ is in that rectangle then:
$||y-x||\leq\sum_{i=1}^n |x_i-y_i|\leq\sum_{i=1}^n\frac{r}{2n}=\frac{r}{2}<r$