Vtyjkyuk

Let $M$ be a smooth manifold. Let $A$ and $B$ be disjoint closed sets of $M$. Show there exists a smooth function $f$ such that $f^-1(0)=A$ and $f^-1(1)=B$.

This is my idea so far,

Since $A$ and $B$ are disjoint closed subsets $M-A,M-B$ is an open cover for $M$. Therefore there exists a partition of unity $psi_1,psi_2$ with $psi_1$ supported in $M-A$ and $psi_2$ supported in $M-B$. Furthermore $psi_1+psi_2=1$. Then $psi_1(1-psi_2)$ is zero on $A$ and $1$ on $B$. I get the inclusions $Asubset f^-1(0)$ and $Bsubset f^-1(1)$. I've tried adding a few more open sets to the cover like $M-(Acup B)$ and $(M-A)-B$ and $(M-B)-A$ but I still can't ensure that the function only vanishes at $A$.

Then I moved on to attempt to find any bump function that is 1 only on $A$ and ran into the same problem.

Am I missing something here or do I just need to be clever with how I define a function. Any hints would be much appreciated. Thanks.

You can construct a proof using the following steps:

• First show that for for an open interval $(a,b)$ there exists a smooth function $phi: mathbb Rto [0,infty)$ such that $phi^-1(0,infty)=(a,b)$, thus $phi$ vanishes outside $(a,b)$. Then do the same for boxes $(a_1,b_1)timesdotstimes (a_n,b_n)$ of $mathbb R^n$, i.e. find $phi:mathbb R^nto [0,infty)$ with $phi^-1(0,infty)=(a_1,b_1)timesdotstimes (a_n,b_n)$


• Show that the above is true when you replace $(a_1,b_1)timesdotstimes (a_n,b_n)$ by an open subset $U$ of $mathbb R^n$. To do this, show that $U$ can be written as union of such open boxes, such that each $pin U$ is contained in finitely many boxes. Then choose a $phi$ as above for each box. The sum of those $phi$'s, will be the desired function for $U$.


• Then prove the same result for the manifold $M$. Use a partition of unity and the previous results for $mathbb R^n$ to do this.


• Finally, if $A,Bsubset M$ are disjoint and closed, then by the previous we can find $phi_1 ,phi_2 :Mto [0,infty)$ such that $phi^-1_1(0)=A$ and $phi^-1_2(0)=B$. The sum $phi_1+phi_2$ never vanishes since $Acap B=emptyset$. Therefore, we can define $phi:= phi_2/(phi_1+phi_2)$. This is smooth on $M$, it attains values in $[0,1]$, and has the desired property that $phi^-1(1)=B, phi^-1(0)=A$