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I encountered the following proposition:

If a function is smooth on an arbitrary set $S\in M$, where $M$ is a smooth manifold, then it has a smooth extension to an open set containing $S$.

It seems the proof needs the partition of unit, but I don't think this proposition is correct.

First of all, I don't know what does smooth on an arbitrary set mean.

Does it mean $f$ is smooth at every point $P$ in $S$, which by definition means there exists a chart $(U, \varphi)$ with $p\in U$ and $f\circ\varphi^{-1}$ if smooth?

If so, does the proposition implies that $S$ is an open set because each point in $S$ should have a neighborhood on which $f$ is defined and smooth?

If so, why emphizie an arbitrary set $S$?

Consider the function $f:\mathbb{Q}\rightarrow\mathbb{R}$ defined as $f(x)=x$ when $x\geq 0$ and $f(x)=-x$ when $x<0$, is it smooth on $\mathbb{Q}$?

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    Oh yes... I didn't notice it.. Thank you!2012-10-22

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