Multidimensional level sets

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Consider a function $mathbf f : mathbb R^n to mathbb R$ as an element of $mathcal C^k(mathbb R^n)$ and a fixed real number $C in mathbb R$. Now define the multidimensional level set $mathcal A_C = displaystylemathbf x in mathbb R^n : mathbf f(mathbf x) = C .$



Does the constrains on $mathbf f$ imply anything for the set $mathcal A_C$?



As an example, consider the simple case of $begincases k=0 \ n=2 endcases$ and the question about the nature of the multidimensional level set $mathcal A_C$ as defined above.



This means that $mathbb f$ maps the plane continuously to the real line. Also assume that there's no ball $B(varepsilon, mathbf x_0) = displaystyle mathbf x - mathbf x_0$ such that $mathbf f$ is constant on $B(varepsilon, mathbf x_0)$. It seems likely that $mathcal A_C$ consists of a countable union – which I stress could be empty – of continuous curves $gamma in mathcal C(mathbb R)$, although a such conjecture obviously have to be proven.



In the more general form, where $k, n in mathbb N$ is chosen arbitrary, can we say something about the nature/topology of the set $mathcal A_C$? Does less restrictive requirements on the function $mathbf f$ enable us to say more, for example if we only required $mathbf f$ to be locally Lipschitz?



(In the more general form, where $mathbf f in mathcal C^k(mathbb R^n)$, it seems like a probable guess that $mathcal A_C$ has to consist of an countable union of sets which are path-connected by $C^k$-curves. Also note that points are, in this sense, curves.)







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  • Well, since $mathbf f$ is continuous then $mathcal A_C$ is closed. I'm not sure how much more you can say without knowing more about $mathbf f$.
    – Theoretical Economist
    Aug 17 at 23:20










  • @TheoreticalEconomist Is this true, though? Consider a constant function $mathbf f(mathbf x) = alpha$ and $mathcal A_alpha = displaystyle mathbf x : mathbf f(mathbf x) = alpha$.. This set is all of $mathbb R^n$ and hence not closed.
    – Markus Klyver
    Aug 17 at 23:31










  • You are mistaken. $mathbb R^n$ is closed.
    – Theoretical Economist
    Aug 17 at 23:34











  • Saying $A_C$ is a union of curves, where points are also curves, isn't saying much. Any set is a union of points.
    – Rahul
    Aug 17 at 23:45















up vote
0
down vote

favorite












Consider a function $mathbf f : mathbb R^n to mathbb R$ as an element of $mathcal C^k(mathbb R^n)$ and a fixed real number $C in mathbb R$. Now define the multidimensional level set $mathcal A_C = displaystylemathbf x in mathbb R^n : mathbf f(mathbf x) = C .$



Does the constrains on $mathbf f$ imply anything for the set $mathcal A_C$?



As an example, consider the simple case of $begincases k=0 \ n=2 endcases$ and the question about the nature of the multidimensional level set $mathcal A_C$ as defined above.



This means that $mathbb f$ maps the plane continuously to the real line. Also assume that there's no ball $B(varepsilon, mathbf x_0) = displaystyle mathbf x - mathbf x_0$ such that $mathbf f$ is constant on $B(varepsilon, mathbf x_0)$. It seems likely that $mathcal A_C$ consists of a countable union – which I stress could be empty – of continuous curves $gamma in mathcal C(mathbb R)$, although a such conjecture obviously have to be proven.



In the more general form, where $k, n in mathbb N$ is chosen arbitrary, can we say something about the nature/topology of the set $mathcal A_C$? Does less restrictive requirements on the function $mathbf f$ enable us to say more, for example if we only required $mathbf f$ to be locally Lipschitz?



(In the more general form, where $mathbf f in mathcal C^k(mathbb R^n)$, it seems like a probable guess that $mathcal A_C$ has to consist of an countable union of sets which are path-connected by $C^k$-curves. Also note that points are, in this sense, curves.)







share|cite|improve this question






















  • Well, since $mathbf f$ is continuous then $mathcal A_C$ is closed. I'm not sure how much more you can say without knowing more about $mathbf f$.
    – Theoretical Economist
    Aug 17 at 23:20










  • @TheoreticalEconomist Is this true, though? Consider a constant function $mathbf f(mathbf x) = alpha$ and $mathcal A_alpha = displaystyle mathbf x : mathbf f(mathbf x) = alpha$.. This set is all of $mathbb R^n$ and hence not closed.
    – Markus Klyver
    Aug 17 at 23:31










  • You are mistaken. $mathbb R^n$ is closed.
    – Theoretical Economist
    Aug 17 at 23:34











  • Saying $A_C$ is a union of curves, where points are also curves, isn't saying much. Any set is a union of points.
    – Rahul
    Aug 17 at 23:45













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Consider a function $mathbf f : mathbb R^n to mathbb R$ as an element of $mathcal C^k(mathbb R^n)$ and a fixed real number $C in mathbb R$. Now define the multidimensional level set $mathcal A_C = displaystylemathbf x in mathbb R^n : mathbf f(mathbf x) = C .$



Does the constrains on $mathbf f$ imply anything for the set $mathcal A_C$?



As an example, consider the simple case of $begincases k=0 \ n=2 endcases$ and the question about the nature of the multidimensional level set $mathcal A_C$ as defined above.



This means that $mathbb f$ maps the plane continuously to the real line. Also assume that there's no ball $B(varepsilon, mathbf x_0) = displaystyle mathbf x - mathbf x_0$ such that $mathbf f$ is constant on $B(varepsilon, mathbf x_0)$. It seems likely that $mathcal A_C$ consists of a countable union – which I stress could be empty – of continuous curves $gamma in mathcal C(mathbb R)$, although a such conjecture obviously have to be proven.



In the more general form, where $k, n in mathbb N$ is chosen arbitrary, can we say something about the nature/topology of the set $mathcal A_C$? Does less restrictive requirements on the function $mathbf f$ enable us to say more, for example if we only required $mathbf f$ to be locally Lipschitz?



(In the more general form, where $mathbf f in mathcal C^k(mathbb R^n)$, it seems like a probable guess that $mathcal A_C$ has to consist of an countable union of sets which are path-connected by $C^k$-curves. Also note that points are, in this sense, curves.)







share|cite|improve this question














Consider a function $mathbf f : mathbb R^n to mathbb R$ as an element of $mathcal C^k(mathbb R^n)$ and a fixed real number $C in mathbb R$. Now define the multidimensional level set $mathcal A_C = displaystylemathbf x in mathbb R^n : mathbf f(mathbf x) = C .$



Does the constrains on $mathbf f$ imply anything for the set $mathcal A_C$?



As an example, consider the simple case of $begincases k=0 \ n=2 endcases$ and the question about the nature of the multidimensional level set $mathcal A_C$ as defined above.



This means that $mathbb f$ maps the plane continuously to the real line. Also assume that there's no ball $B(varepsilon, mathbf x_0) = displaystyle mathbf x - mathbf x_0$ such that $mathbf f$ is constant on $B(varepsilon, mathbf x_0)$. It seems likely that $mathcal A_C$ consists of a countable union – which I stress could be empty – of continuous curves $gamma in mathcal C(mathbb R)$, although a such conjecture obviously have to be proven.



In the more general form, where $k, n in mathbb N$ is chosen arbitrary, can we say something about the nature/topology of the set $mathcal A_C$? Does less restrictive requirements on the function $mathbf f$ enable us to say more, for example if we only required $mathbf f$ to be locally Lipschitz?



(In the more general form, where $mathbf f in mathcal C^k(mathbb R^n)$, it seems like a probable guess that $mathcal A_C$ has to consist of an countable union of sets which are path-connected by $C^k$-curves. Also note that points are, in this sense, curves.)









share|cite|improve this question













share|cite|improve this question




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edited Aug 17 at 23:54

























asked Aug 17 at 23:05









Markus Klyver

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  • Well, since $mathbf f$ is continuous then $mathcal A_C$ is closed. I'm not sure how much more you can say without knowing more about $mathbf f$.
    – Theoretical Economist
    Aug 17 at 23:20










  • @TheoreticalEconomist Is this true, though? Consider a constant function $mathbf f(mathbf x) = alpha$ and $mathcal A_alpha = displaystyle mathbf x : mathbf f(mathbf x) = alpha$.. This set is all of $mathbb R^n$ and hence not closed.
    – Markus Klyver
    Aug 17 at 23:31










  • You are mistaken. $mathbb R^n$ is closed.
    – Theoretical Economist
    Aug 17 at 23:34











  • Saying $A_C$ is a union of curves, where points are also curves, isn't saying much. Any set is a union of points.
    – Rahul
    Aug 17 at 23:45

















  • Well, since $mathbf f$ is continuous then $mathcal A_C$ is closed. I'm not sure how much more you can say without knowing more about $mathbf f$.
    – Theoretical Economist
    Aug 17 at 23:20










  • @TheoreticalEconomist Is this true, though? Consider a constant function $mathbf f(mathbf x) = alpha$ and $mathcal A_alpha = displaystyle mathbf x : mathbf f(mathbf x) = alpha$.. This set is all of $mathbb R^n$ and hence not closed.
    – Markus Klyver
    Aug 17 at 23:31










  • You are mistaken. $mathbb R^n$ is closed.
    – Theoretical Economist
    Aug 17 at 23:34











  • Saying $A_C$ is a union of curves, where points are also curves, isn't saying much. Any set is a union of points.
    – Rahul
    Aug 17 at 23:45
















Well, since $mathbf f$ is continuous then $mathcal A_C$ is closed. I'm not sure how much more you can say without knowing more about $mathbf f$.
– Theoretical Economist
Aug 17 at 23:20




Well, since $mathbf f$ is continuous then $mathcal A_C$ is closed. I'm not sure how much more you can say without knowing more about $mathbf f$.
– Theoretical Economist
Aug 17 at 23:20












@TheoreticalEconomist Is this true, though? Consider a constant function $mathbf f(mathbf x) = alpha$ and $mathcal A_alpha = displaystyle mathbf x : mathbf f(mathbf x) = alpha$.. This set is all of $mathbb R^n$ and hence not closed.
– Markus Klyver
Aug 17 at 23:31




@TheoreticalEconomist Is this true, though? Consider a constant function $mathbf f(mathbf x) = alpha$ and $mathcal A_alpha = displaystyle mathbf x : mathbf f(mathbf x) = alpha$.. This set is all of $mathbb R^n$ and hence not closed.
– Markus Klyver
Aug 17 at 23:31












You are mistaken. $mathbb R^n$ is closed.
– Theoretical Economist
Aug 17 at 23:34





You are mistaken. $mathbb R^n$ is closed.
– Theoretical Economist
Aug 17 at 23:34













Saying $A_C$ is a union of curves, where points are also curves, isn't saying much. Any set is a union of points.
– Rahul
Aug 17 at 23:45





Saying $A_C$ is a union of curves, where points are also curves, isn't saying much. Any set is a union of points.
– Rahul
Aug 17 at 23:45
















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