Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm really sorry if the title is not at all descriptive but I really cannot come up with anything better.

Let $\mathcal{A} \subset \mathbb{R}^{\mathbb{R}^{n}}$ be nonempty and $\mathcal{S(\mathcal{A})}$ be a family of all sets of the form $$\bigcup_{i=1}^{p} \bigcap_{j=1}^{q} \{ x\in \mathbb{R}^{n}: f_{i}(x)=0, \ g_{ij}(x)>0 \} \ \ p,q \in \mathbb{N}, \ f_{i}, g_{ij} \in \mathcal{A}\;.$$ Show that finite unions, finite intersections and complements (here some assumptions about $\mathcal A$ may be needed) of $\mathcal{S(\mathcal{A})}$'s elements are also members of $\mathcal{S(\mathcal{A})}$.

We could consider $\mathcal{A}$ as a set of functions which to every polynomial $P \in \mathbb{R}^n$ assign $x \in \mathbb{R}$. Could that $x$ be a root? I'm not sure. It doesn't make much sense that way.

Could you help me?

share|cite|improve this question
up vote 1 down vote accepted

First note that if $g_{i,q+1}:\Bbb R^n\to\Bbb R$ is any constant function with positive value,

$$\bigcap_{j=1}^q\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}=\bigcap_{j=1}^{q+1}\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}\;,$$


$$\begin{align*} \left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{i,q+1}(x)>0\right\}&=\left\{x\in\Bbb R^n:f_i(x)=0\right\}\\ &\supseteq\bigcap_{j=1}^q\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}\;. \end{align*}$$

Suppose that

$$S=\bigcup_{i=1}^p\bigcap_{j=1}^q\left\{x\in\Bbb R^n:f_{i}(x)=0\text{ and }g_{ij}(x)>0\right\}$$


$$T=\bigcup_{i=1}^r\bigcap_{j=1}^s\left\{x\in\Bbb R^n:h_{i}(x)=0\text{ and }k_{ij}(x)>0\right\}\;.$$

Without loss of generality assume that $q\le s$. If $q<s$, let $g_{ij}$ be a positive constant function for each $i=1,\dots,p$ and $j=q+1,\dots,s$; by the previous observation we may then write

$$S=\bigcup_{i=1}^p\bigcap_{j=1}^s\left\{x\in\Bbb R^n:f_{i}(x)=0\text{ and }g_{ij}(x)>0\right\}\;.$$

Now for $i=1,\dots,r$ and $j=1,\dots,s$ let $f_{p+i}=h_i$ and $g_{p+i,j}=k_{ij}$; then

$$S\cup T=\bigcup_{i=1}^{p+r}\bigcap_{j=1}^s\left\{x\in\Bbb R^n:f_{i}(x)=0\text{ and }g_{ij}(x)>0\right\}\in\mathcal{S}\;,$$

and an easy induction shows that $\mathcal{S}$ is closed under finite unions.

To show that $T\cap S\in\mathcal{S}$, note first that

$$\begin{align*} &T\cap\bigcup_{i=1}^p\bigcap_{j=1}^q\left\{x\in\Bbb R^n:f_{i}(x)=0\text{ and }g_{ij}(x)>0\right\}\\ &\qquad\quad=\bigcup_{i=1}^p\left(T\cap\bigcap_{j=1}^q\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}\right)\;, \end{align*}$$

so it suffices to show that

$$T\cap\bigcap_{j=1}^q\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}\in\mathcal{S}$$

for $i=1,\dots,p$. This in turn will follow by an easy induction if we can show that

$$T\cap\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}\in\mathcal{S}$$

for a fixed $i$ and $j$. But

$$\begin{align*} &T\cap\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}\\ &\qquad=\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}\cap\bigcup_{i=1}^r\bigcap_{j=1}^s\left\{x\in\Bbb R^n:h_{i}(x)=0\text{ and }k_{ij}(x)>0\right\} \end{align*}$$

is a finite union of sets of the form

$$\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}\cap\bigcap_{j=1}^s\left\{x\in\Bbb R^n:h_{i}(x)=0\text{ and }k_{ij}(x)>0\right\}\;,$$

which clearly belong to $\mathcal{S}$, so indeed

$$T\cap\left\{x\in\Bbb R^n:f_i(x)=0\text{ and }g_{ij}(x)>0\right\}\in\mathcal{S}$$

and hence $T\cap S\in\mathcal{S}$.

$$\begin{align*} \Bbb R^n\setminus S&=\Bbb R^n\setminus\bigcup_{i=1}^p\bigcap_{j=1}^q\left\{x\in\Bbb R^n:f_{i}(x)=0\text{ and }g_{ij}(x)>0\right\}\\ &=\bigcap_{i=1}^p\bigcup_{j=1}^q\left\{x\in\Bbb R^n:f_i(x)\ne 0,\text{ or }g_{ij}(x)\le 0\text{ for some }j\right\}\;, \end{align*}$$

so by what we’ve already done it suffices to show that each

$$\left\{x\in\Bbb R^n:f_i(x)\ne 0,\text{ or }g_{ij}(x)\le 0\text{ for some }j\right\}\in\mathcal{S}\;.$$

The set $$\left\{x\in\Bbb R^n:f_i(x)\ne 0,\text{ or }g_{ij}(x)\le 0\text{ for some }j\right\}$$

is the union of the following sets:

$$\begin{align*} &\left\{x\in\Bbb R^n:f_i(x)>0\right\}\;,\\ &\left\{x\in\Bbb R^n:f_i(x)<0\right\}=\left\{x\in\Bbb R^n:-f_i(x)>0\right\}\;,\\ &\left\{x\in\Bbb R^n:g_{ij}(x)=0\right\}\;,\text{ and}\\ &\left\{x\in\Bbb R^n:g_{ij}(x)<0\right\}=\left\{x\in\Bbb R^n:-g_{ij}(x)<0\right\}\;. \end{align*}$$

Can you show that each of them is in $\mathcal{S}$ and hence that their union is as well?

share|cite|improve this answer

The elements of $\mathbb{R}^{\mathbb{R}^n}$ are the functions from $\mathbb{R}^n$ to $\mathbb{R}$, and so $\mathcal{A}$ is some set of such functions. That is, if $f\in\mathcal{A}$, then $f$ assigns to each point $p\in\mathbb{R}^n$ a number $f(p)\in\mathbb{R}$.

Note that these don't have to be polynomial functions necessarily; for example, the function $f$ defined by $f(x,y,z) = |x|^{y+\sin(z)}$ is in $\mathbb{R}^{\mathbb{R}^3}$ but is not a polynomial function.

When you wrote $x$ rather than $f(p)$ in your question, note that it is not playing the same role as the variable $x$ in definition of $\mathcal{S}(\mathcal{A})$ -- the $x$ involved there are elements of $\mathbb{R}^n$, not $\mathbb{R}$. If $n>1$ then it doesn't actually make sense for $f(p)$ to be a "root" of $f$, because $f(f(p))$ is not defined.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.