Prove that the Cartesian product of the open intervals $$(a_1, b_1)\times(a_2, b_2)\times\cdots\times(a_{n-1}, b_{n-1})\times(a_n, b_n)$$ is an open set in $\mathbb R^n$.
Let us chose any point say $x = (x_1, x_2,\ldots,x_n)$ from the set $$(a_1, b_1)\times(a_2, b_2)\times\cdots\times(a_{n-1}, b_{n-1})\times(a_n, b_n)$$ then for each $k=1, 2,\ldots,n$, we see that $x_k$ belongs to $(a_k, b_k)$. Now as $k$ is finite so chose $$r < \text{min}(|x_k-a_k|, |x_k-b_k|),$$ where $k=1, 2,\ldots,n$, then if we choose any point in this $n$-ball of the point $x$ with radius $r$ then we have \begin{align*}\text{min}(|x_k-a_k|, |x_k-b_k|)> r &> |y-x|\\&=\sqrt{((y_1-x_1)^2+(y_2-x_2)^2+\cdots+(y_n-x_n)^2)} \geq |y_k-x_k|\end{align*} for any $k=1, 2,\ldots,n$, and hence each $y_k$ lies in the interval $(a_k, b_k)$. And hence each such point in this n-ball of the point $x$ lies in the set $$(a_1, b_1)\times(a_2, b_2)\times\cdots\times(a_{n-1}, b_{n-1})\times(a_n, b_n)$$ and as $x$ was an arbitrary point, so the given set is an open set.
Is My Proof Correct??
