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Consider functions $f_i(x)$ that map $\mathbb{R}^n$ to $\mathbb{R}$ for $i\in{1,\dots,k}$. Assume these functions are monotonically increasing in their arguments and continuous everywhere Also, one can assume that they are bounded. Consider the set of inequalities $f_i(x)\leq 0$. What are some properties of the feasible set of this set of inequalities defined as $$\mathcal{F}=\bigcap_{1\leq i\leq k}\{x~|~f_i(x)\leq 0,~\forall i\}$$

For instance, consider the case of only one function ($k=1$). Let $x=(x_1,\dots,x_n)$ be a feasible point such that $f_1(x)$. Then define the set $$\mathcal{S}_x=\{y~|~y_j\leq x_j\}$$ Essentially $\mathcal{S}_x$ is set of all vectors that are component-wise lesser than $x$. This is a convex set. Moreover, due to monotonicity of $f_1(.)$, all points in $S_x$ are feasible as well, i.e. $\mathcal{S}_x\subset\mathcal{F}$. In fact, we can define the set $$\mathcal{A}=\{x~|~f_1(x)=0\}$$ It is easy to see that the set $$\mathcal{B}=\bigcup_{x\in \mathcal{A}}\mathcal{S}_x$$ is also a feasible set ($\mathcal{B}\subset\mathcal{F}$). Due to monotonicity of $f_1(.)$, we can prove that $\mathcal{B}$ is the only feasible set ($\mathcal{F}\subset\mathcal{B}$) . However, I am not able to generalize this kind of arguments beyond one function ($k>1$). Appreciate any help in this direction

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  • $\begingroup$ it's not quite clear what is the question: will you state it more precisely? Also, does "the feasible set of this set of inequalities" mean the set $\cap_{1\le i \le n}\{x\in\mathbb{R}^n : f_i(x)\le0\}$? $\endgroup$ – Pietro Majer May 17 '18 at 9:00
  • $\begingroup$ yes it does. Updated the question to reflect that. $\endgroup$ – dineshdileep May 18 '18 at 6:06
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You can reduce the general case to the case of one function by considering $f:=\max_1^k f_i$. Then $\mathcal F=\{x\colon f(x)\le0\}$ and $f$ is increasing in its arguments and continuous everywhere.

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