# A result attributed to Whitney

One of the basic results of real analysis says that any closed subset of a smooth ($C^\infty$) manifold $M$ is the set of zeros of some map $\lambda\in C^\infty(M;[0,1])$. This result (or some equivalent variations of it) is usually attributed to Hassler Whitney.

Question: exactly where (title of Whitney's paper and page) is this stated?

• – Pietro Majer Feb 10 '15 at 19:00
• @Pietro: following your link, Peter Michor says:" Maybe, this result is hidden in Whitney, Hassler: Analytic extensions of differentiable functions defined in closed sets, Trans. AMS 36 (1934), 63--89." But where? – Nautilus Feb 10 '15 at 19:10
• I don't have Whitney's paper in front of me right now, but IIRC the extension operator constructed by Whitney when applied to the zero function (defined on a closed set $F$) has the property that the constructed function is comparable to the distance function from $F$. Do you need something else? – Willie Wong Feb 11 '15 at 3:13
• This is actually a very good elementary exercise. See Robert Israel's answer for $\mathbb R^n$, and glue by partition of unity: math.stackexchange.com/questions/791248/…. Whitney was concerned with much more subtle issues. – Igor Belegradek May 12 '15 at 0:19
• If you really like using Whitney's theorem, here is how (I will only do the case $M=\mathbb R^n$ here): Whitney proves that there is a function $f$ that vanishes on the given closed set $A$ and is analytic elsewhere. Then $f^2$ have the same property. Since zeros of analytic maps do not accumulate, they are isolated in the complement of $A$. Add a bump function to $f^2$ near every zero. This gives the desired function that vanishes precisely on $A$. – Igor Belegradek May 12 '15 at 0:47

Analytic extensions of differentiable functions defined in closed sets

see this discussion

Actually, Whitney's Theorem is stronger in two ways. On the one hand, every $C^m$-function $F:K\rightarrow{\mathbb R}$ ($K$ the closed set) can be matched at order $m$ by a $C^m$-function $W(F):M\rightarrow{\mathbb R}$ which is $C^\infty$ away from $K$. On the other hand, the operator $W$ can be chosen linear and continuous (for the $C^m$-norms).