A function that separates points of $mathbb{R}^N$












1












$begingroup$


Let $K$ be a compact non-empty subset of $mathbb{R}^N$. If $x in mathbb{R}^N$ and if $(k_1, k_2, dots, k_N), k_i in mathbb{Z}_+$, consider the function
$$x mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$



I want to show that the set $mathcal{A}$ of all linear combinations of such functions separates points of $K$.




Definition: a set of functions $S$ from a set $D$ to a set $C$ is said to separate the points of $D$ if for any two distinct
elements $x$ and $y$ of $D$, there exists a function $f$ in $S$ so that $f(x) neq f(y)$.




If $K subset mathbb{R}$, it's easy since the polynomial $f(x)=x$ separates all points, but I'm jammed with $K subset mathbb{R}^N$. The sum of coordinates to rational exponents was an easy start, but consider such a function $f$ and take $x,y in mathbb{R}^N, x = (1, 0, dots, 0), y = (0, 1, dots, 0)$, then $f(x) = f(y) = 1$. What could that function be?






real-analysis polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    The function $f(x_1,x_2) = x_1 + 2x_2$ separates $(1,0)$ and $(0,1)$.
    $endgroup$
    – mlerma54
    Dec 5 '18 at 15:47










  • $begingroup$
    I presume the condition on the $k_i$ is that they are non-negative integers, correct?
    $endgroup$
    – mlerma54
    Dec 5 '18 at 15:48










  • $begingroup$
    @mlerma54: the condition on the $k_i$ you stated is correct.
    $endgroup$
    – user71487
    Dec 5 '18 at 15:52






  • 1




    $begingroup$
    The solution I had in mind is $f_i(x_1,dots,x_n) = x_i$, which separates points that differ in the $i$-th coordinate, so $S={f_1,dots,f_n}$ would work.
    $endgroup$
    – mlerma54
    Dec 5 '18 at 16:22










  • $begingroup$
    @mlerma54: True. If you post the answer, I'll choose it just for posterity's sake.
    $endgroup$
    – user71487
    Dec 5 '18 at 16:43
















1












$begingroup$


Let $K$ be a compact non-empty subset of $mathbb{R}^N$. If $x in mathbb{R}^N$ and if $(k_1, k_2, dots, k_N), k_i in mathbb{Z}_+$, consider the function
$$x mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$



I want to show that the set $mathcal{A}$ of all linear combinations of such functions separates points of $K$.




Definition: a set of functions $S$ from a set $D$ to a set $C$ is said to separate the points of $D$ if for any two distinct
elements $x$ and $y$ of $D$, there exists a function $f$ in $S$ so that $f(x) neq f(y)$.




If $K subset mathbb{R}$, it's easy since the polynomial $f(x)=x$ separates all points, but I'm jammed with $K subset mathbb{R}^N$. The sum of coordinates to rational exponents was an easy start, but consider such a function $f$ and take $x,y in mathbb{R}^N, x = (1, 0, dots, 0), y = (0, 1, dots, 0)$, then $f(x) = f(y) = 1$. What could that function be?






real-analysis polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    The function $f(x_1,x_2) = x_1 + 2x_2$ separates $(1,0)$ and $(0,1)$.
    $endgroup$
    – mlerma54
    Dec 5 '18 at 15:47










  • $begingroup$
    I presume the condition on the $k_i$ is that they are non-negative integers, correct?
    $endgroup$
    – mlerma54
    Dec 5 '18 at 15:48










  • $begingroup$
    @mlerma54: the condition on the $k_i$ you stated is correct.
    $endgroup$
    – user71487
    Dec 5 '18 at 15:52






  • 1




    $begingroup$
    The solution I had in mind is $f_i(x_1,dots,x_n) = x_i$, which separates points that differ in the $i$-th coordinate, so $S={f_1,dots,f_n}$ would work.
    $endgroup$
    – mlerma54
    Dec 5 '18 at 16:22










  • $begingroup$
    @mlerma54: True. If you post the answer, I'll choose it just for posterity's sake.
    $endgroup$
    – user71487
    Dec 5 '18 at 16:43














1












1








1





$begingroup$


Let $K$ be a compact non-empty subset of $mathbb{R}^N$. If $x in mathbb{R}^N$ and if $(k_1, k_2, dots, k_N), k_i in mathbb{Z}_+$, consider the function
$$x mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$



I want to show that the set $mathcal{A}$ of all linear combinations of such functions separates points of $K$.




Definition: a set of functions $S$ from a set $D$ to a set $C$ is said to separate the points of $D$ if for any two distinct
elements $x$ and $y$ of $D$, there exists a function $f$ in $S$ so that $f(x) neq f(y)$.




If $K subset mathbb{R}$, it's easy since the polynomial $f(x)=x$ separates all points, but I'm jammed with $K subset mathbb{R}^N$. The sum of coordinates to rational exponents was an easy start, but consider such a function $f$ and take $x,y in mathbb{R}^N, x = (1, 0, dots, 0), y = (0, 1, dots, 0)$, then $f(x) = f(y) = 1$. What could that function be?






real-analysis polynomials







share|cite|improve this question









$endgroup$




Let $K$ be a compact non-empty subset of $mathbb{R}^N$. If $x in mathbb{R}^N$ and if $(k_1, k_2, dots, k_N), k_i in mathbb{Z}_+$, consider the function
$$x mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$



I want to show that the set $mathcal{A}$ of all linear combinations of such functions separates points of $K$.




Definition: a set of functions $S$ from a set $D$ to a set $C$ is said to separate the points of $D$ if for any two distinct
elements $x$ and $y$ of $D$, there exists a function $f$ in $S$ so that $f(x) neq f(y)$.




If $K subset mathbb{R}$, it's easy since the polynomial $f(x)=x$ separates all points, but I'm jammed with $K subset mathbb{R}^N$. The sum of coordinates to rational exponents was an easy start, but consider such a function $f$ and take $x,y in mathbb{R}^N, x = (1, 0, dots, 0), y = (0, 1, dots, 0)$, then $f(x) = f(y) = 1$. What could that function be?






real-analysis polynomials




real-analysis polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 15:01









user71487user71487

948




948












  • $begingroup$
    The function $f(x_1,x_2) = x_1 + 2x_2$ separates $(1,0)$ and $(0,1)$.
    $endgroup$
    – mlerma54
    Dec 5 '18 at 15:47










  • $begingroup$
    I presume the condition on the $k_i$ is that they are non-negative integers, correct?
    $endgroup$
    – mlerma54
    Dec 5 '18 at 15:48










  • $begingroup$
    @mlerma54: the condition on the $k_i$ you stated is correct.
    $endgroup$
    – user71487
    Dec 5 '18 at 15:52






  • 1




    $begingroup$
    The solution I had in mind is $f_i(x_1,dots,x_n) = x_i$, which separates points that differ in the $i$-th coordinate, so $S={f_1,dots,f_n}$ would work.
    $endgroup$
    – mlerma54
    Dec 5 '18 at 16:22










  • $begingroup$
    @mlerma54: True. If you post the answer, I'll choose it just for posterity's sake.
    $endgroup$
    – user71487
    Dec 5 '18 at 16:43


















  • $begingroup$
    The function $f(x_1,x_2) = x_1 + 2x_2$ separates $(1,0)$ and $(0,1)$.
    $endgroup$
    – mlerma54
    Dec 5 '18 at 15:47










  • $begingroup$
    I presume the condition on the $k_i$ is that they are non-negative integers, correct?
    $endgroup$
    – mlerma54
    Dec 5 '18 at 15:48










  • $begingroup$
    @mlerma54: the condition on the $k_i$ you stated is correct.
    $endgroup$
    – user71487
    Dec 5 '18 at 15:52






  • 1




    $begingroup$
    The solution I had in mind is $f_i(x_1,dots,x_n) = x_i$, which separates points that differ in the $i$-th coordinate, so $S={f_1,dots,f_n}$ would work.
    $endgroup$
    – mlerma54
    Dec 5 '18 at 16:22










  • $begingroup$
    @mlerma54: True. If you post the answer, I'll choose it just for posterity's sake.
    $endgroup$
    – user71487
    Dec 5 '18 at 16:43
















$begingroup$
The function $f(x_1,x_2) = x_1 + 2x_2$ separates $(1,0)$ and $(0,1)$.
$endgroup$
– mlerma54
Dec 5 '18 at 15:47




$begingroup$
The function $f(x_1,x_2) = x_1 + 2x_2$ separates $(1,0)$ and $(0,1)$.
$endgroup$
– mlerma54
Dec 5 '18 at 15:47












$begingroup$
I presume the condition on the $k_i$ is that they are non-negative integers, correct?
$endgroup$
– mlerma54
Dec 5 '18 at 15:48




$begingroup$
I presume the condition on the $k_i$ is that they are non-negative integers, correct?
$endgroup$
– mlerma54
Dec 5 '18 at 15:48












$begingroup$
@mlerma54: the condition on the $k_i$ you stated is correct.
$endgroup$
– user71487
Dec 5 '18 at 15:52




$begingroup$
@mlerma54: the condition on the $k_i$ you stated is correct.
$endgroup$
– user71487
Dec 5 '18 at 15:52




1




1




$begingroup$
The solution I had in mind is $f_i(x_1,dots,x_n) = x_i$, which separates points that differ in the $i$-th coordinate, so $S={f_1,dots,f_n}$ would work.
$endgroup$
– mlerma54
Dec 5 '18 at 16:22




$begingroup$
The solution I had in mind is $f_i(x_1,dots,x_n) = x_i$, which separates points that differ in the $i$-th coordinate, so $S={f_1,dots,f_n}$ would work.
$endgroup$
– mlerma54
Dec 5 '18 at 16:22












$begingroup$
@mlerma54: True. If you post the answer, I'll choose it just for posterity's sake.
$endgroup$
– user71487
Dec 5 '18 at 16:43




$begingroup$
@mlerma54: True. If you post the answer, I'll choose it just for posterity's sake.
$endgroup$
– user71487
Dec 5 '18 at 16:43










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$begingroup$

We have that $f_i(x_1,dots,x_n) = x_i$ separates points that differ in the $i$-th coordinate, so $S = {f_1,dots,f_n}$ solves the problem.






share|cite|improve this answer









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    $begingroup$

    We have that $f_i(x_1,dots,x_n) = x_i$ separates points that differ in the $i$-th coordinate, so $S = {f_1,dots,f_n}$ solves the problem.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      We have that $f_i(x_1,dots,x_n) = x_i$ separates points that differ in the $i$-th coordinate, so $S = {f_1,dots,f_n}$ solves the problem.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        We have that $f_i(x_1,dots,x_n) = x_i$ separates points that differ in the $i$-th coordinate, so $S = {f_1,dots,f_n}$ solves the problem.






        share|cite|improve this answer









        $endgroup$



        We have that $f_i(x_1,dots,x_n) = x_i$ separates points that differ in the $i$-th coordinate, so $S = {f_1,dots,f_n}$ solves the problem.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 16:51









        mlerma54mlerma54

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