If $Csubsetmathbb{R}^n$ is closed, then $C=f^{-1}(0)$ for some smooth $f:mathbb{R}^ntomathbb{R}$.












3












$begingroup$



Let $Csubset mathbb{R}^n$ closed. Prove that there is a smooth function $f:mathbb{R}^ntomathbb{R}$ such that $C=f^{-1}(0)$.




I've found this solution in the internet: take a cover of balls ${B_i}_i$ of $mathbb{R}^nsetminus C$ and a partition of unity ${f_i}_i$ subordinate to ${B_i}_i$ and define $f:=sum_{i=1}^infty frac{f_i}{2^iM_i}$, where:
$$M_i:=supleft{frac{partial^kf_i}{partial x_{j_1}cdots partial x_{j_k}}(p)mid 1leq j_1<...<j_kleq n,, kleq i,,pin B_iright}$$



My question is: why do we need to define these $M_i$'s?



Why can't we just define $f=sum_{i=1}^infty frac{f_i}{2^i}$ for example?



In that case, since ${B_i}_i$ can be considered locally finite, then for $pinmathbb{R}^n$, there is some open $Usubsetmathbb{R}^nsetminus C$ containing $p$ such that $f_i|_Uequiv 0$ for all but finitely many $i$ (say $i_1,...,i_k$), so $f|_U=frac{f_{i_1}}{2^{i_1}}+...+frac{f_{i_k}}{2^{i_k}}$, which is obviously smooth.



What am I missing?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
    $endgroup$
    – Tim kinsella
    Feb 2 at 19:05












  • $begingroup$
    You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
    $endgroup$
    – rmdmc89
    Feb 2 at 19:38










  • $begingroup$
    if $pin C$ then there is no such $U$
    $endgroup$
    – Tim kinsella
    Feb 2 at 22:53






  • 1




    $begingroup$
    my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
    $endgroup$
    – Tim kinsella
    Feb 2 at 22:55










  • $begingroup$
    Oh, I see what you mean. I'll think about it
    $endgroup$
    – rmdmc89
    Feb 3 at 12:04
















3












$begingroup$



Let $Csubset mathbb{R}^n$ closed. Prove that there is a smooth function $f:mathbb{R}^ntomathbb{R}$ such that $C=f^{-1}(0)$.




I've found this solution in the internet: take a cover of balls ${B_i}_i$ of $mathbb{R}^nsetminus C$ and a partition of unity ${f_i}_i$ subordinate to ${B_i}_i$ and define $f:=sum_{i=1}^infty frac{f_i}{2^iM_i}$, where:
$$M_i:=supleft{frac{partial^kf_i}{partial x_{j_1}cdots partial x_{j_k}}(p)mid 1leq j_1<...<j_kleq n,, kleq i,,pin B_iright}$$



My question is: why do we need to define these $M_i$'s?



Why can't we just define $f=sum_{i=1}^infty frac{f_i}{2^i}$ for example?



In that case, since ${B_i}_i$ can be considered locally finite, then for $pinmathbb{R}^n$, there is some open $Usubsetmathbb{R}^nsetminus C$ containing $p$ such that $f_i|_Uequiv 0$ for all but finitely many $i$ (say $i_1,...,i_k$), so $f|_U=frac{f_{i_1}}{2^{i_1}}+...+frac{f_{i_k}}{2^{i_k}}$, which is obviously smooth.



What am I missing?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
    $endgroup$
    – Tim kinsella
    Feb 2 at 19:05












  • $begingroup$
    You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
    $endgroup$
    – rmdmc89
    Feb 2 at 19:38










  • $begingroup$
    if $pin C$ then there is no such $U$
    $endgroup$
    – Tim kinsella
    Feb 2 at 22:53






  • 1




    $begingroup$
    my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
    $endgroup$
    – Tim kinsella
    Feb 2 at 22:55










  • $begingroup$
    Oh, I see what you mean. I'll think about it
    $endgroup$
    – rmdmc89
    Feb 3 at 12:04














3












3








3


1



$begingroup$



Let $Csubset mathbb{R}^n$ closed. Prove that there is a smooth function $f:mathbb{R}^ntomathbb{R}$ such that $C=f^{-1}(0)$.




I've found this solution in the internet: take a cover of balls ${B_i}_i$ of $mathbb{R}^nsetminus C$ and a partition of unity ${f_i}_i$ subordinate to ${B_i}_i$ and define $f:=sum_{i=1}^infty frac{f_i}{2^iM_i}$, where:
$$M_i:=supleft{frac{partial^kf_i}{partial x_{j_1}cdots partial x_{j_k}}(p)mid 1leq j_1<...<j_kleq n,, kleq i,,pin B_iright}$$



My question is: why do we need to define these $M_i$'s?



Why can't we just define $f=sum_{i=1}^infty frac{f_i}{2^i}$ for example?



In that case, since ${B_i}_i$ can be considered locally finite, then for $pinmathbb{R}^n$, there is some open $Usubsetmathbb{R}^nsetminus C$ containing $p$ such that $f_i|_Uequiv 0$ for all but finitely many $i$ (say $i_1,...,i_k$), so $f|_U=frac{f_{i_1}}{2^{i_1}}+...+frac{f_{i_k}}{2^{i_k}}$, which is obviously smooth.



What am I missing?










share|cite|improve this question











$endgroup$





Let $Csubset mathbb{R}^n$ closed. Prove that there is a smooth function $f:mathbb{R}^ntomathbb{R}$ such that $C=f^{-1}(0)$.




I've found this solution in the internet: take a cover of balls ${B_i}_i$ of $mathbb{R}^nsetminus C$ and a partition of unity ${f_i}_i$ subordinate to ${B_i}_i$ and define $f:=sum_{i=1}^infty frac{f_i}{2^iM_i}$, where:
$$M_i:=supleft{frac{partial^kf_i}{partial x_{j_1}cdots partial x_{j_k}}(p)mid 1leq j_1<...<j_kleq n,, kleq i,,pin B_iright}$$



My question is: why do we need to define these $M_i$'s?



Why can't we just define $f=sum_{i=1}^infty frac{f_i}{2^i}$ for example?



In that case, since ${B_i}_i$ can be considered locally finite, then for $pinmathbb{R}^n$, there is some open $Usubsetmathbb{R}^nsetminus C$ containing $p$ such that $f_i|_Uequiv 0$ for all but finitely many $i$ (say $i_1,...,i_k$), so $f|_U=frac{f_{i_1}}{2^{i_1}}+...+frac{f_{i_k}}{2^{i_k}}$, which is obviously smooth.



What am I missing?







real-analysis smooth-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 2 at 19:36







rmdmc89

















asked Feb 2 at 17:16









rmdmc89rmdmc89

2,2421923




2,2421923








  • 1




    $begingroup$
    i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
    $endgroup$
    – Tim kinsella
    Feb 2 at 19:05












  • $begingroup$
    You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
    $endgroup$
    – rmdmc89
    Feb 2 at 19:38










  • $begingroup$
    if $pin C$ then there is no such $U$
    $endgroup$
    – Tim kinsella
    Feb 2 at 22:53






  • 1




    $begingroup$
    my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
    $endgroup$
    – Tim kinsella
    Feb 2 at 22:55










  • $begingroup$
    Oh, I see what you mean. I'll think about it
    $endgroup$
    – rmdmc89
    Feb 3 at 12:04














  • 1




    $begingroup$
    i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
    $endgroup$
    – Tim kinsella
    Feb 2 at 19:05












  • $begingroup$
    You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
    $endgroup$
    – rmdmc89
    Feb 2 at 19:38










  • $begingroup$
    if $pin C$ then there is no such $U$
    $endgroup$
    – Tim kinsella
    Feb 2 at 22:53






  • 1




    $begingroup$
    my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
    $endgroup$
    – Tim kinsella
    Feb 2 at 22:55










  • $begingroup$
    Oh, I see what you mean. I'll think about it
    $endgroup$
    – rmdmc89
    Feb 3 at 12:04








1




1




$begingroup$
i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
$endgroup$
– Tim kinsella
Feb 2 at 19:05






$begingroup$
i'm not certain, but what if $pin C$ in your last paragraph? the $B_i$ are only locally finite at points outside of C, right?
$endgroup$
– Tim kinsella
Feb 2 at 19:05














$begingroup$
You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
$endgroup$
– rmdmc89
Feb 2 at 19:38




$begingroup$
You're right @Timkinsella, I should have said $Usubset mathbb{R}^nsetminus C$. I've corrected it
$endgroup$
– rmdmc89
Feb 2 at 19:38












$begingroup$
if $pin C$ then there is no such $U$
$endgroup$
– Tim kinsella
Feb 2 at 22:53




$begingroup$
if $pin C$ then there is no such $U$
$endgroup$
– Tim kinsella
Feb 2 at 22:53




1




1




$begingroup$
my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
$endgroup$
– Tim kinsella
Feb 2 at 22:55




$begingroup$
my point (of which, as i say, i'm not certain) is that the $M_i$ are there to guarantee smoothness at points of $C$ somehow.
$endgroup$
– Tim kinsella
Feb 2 at 22:55












$begingroup$
Oh, I see what you mean. I'll think about it
$endgroup$
– rmdmc89
Feb 3 at 12:04




$begingroup$
Oh, I see what you mean. I'll think about it
$endgroup$
– rmdmc89
Feb 3 at 12:04










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