I am facing a problem as follows. Given a set $A = \{X \mid X \in R^n, 0 \le f(X) \le \xi\}$, where $X=[x_1,x_2,\dots,x_n]^\top, \xi > 0$, $f:R^n \rightarrow R$ is a continuous function. Can we prove that there exist some numbers $a_1, a_2,\dots, a_n$ and $c_1,x_2,\dots,c_n$ such that \begin{align} &|x_1 - a_1| \le c_1 \xi\\ &|x_2 - a_2| \le c_2 \xi\\ &\dots\\ &|x_n - a_n| \le c_n \xi\\ \end{align} which is the boundary of a hypercube. For example, consider a simple case where $f$ is the fuction of a circle $0 \le f(x, y) = x^2 + y^2 \le 1$, we can easily verify that \begin{align} |x - 0 | \le 1/\sqrt{2} \times 1\\ |y - 0 | \le 1/\sqrt{2} \times 1\\ \end{align} is the square inscribed in this circles.
I don't know where to start with this problem.
If $f$ is continuous and $B = \{X \in \mathbb R^n \mid 0 < f(X) < \xi \}$ not empty then $B $ is open. As all the norms in $\mathbb R^n$ are equivalent, $B$ will contain an open hypercube $H_x$ containing $x$ for all $x \in B$.