Consider a nonempty subset $A$ of a smooth manifold $M$ and suppose that every smooth function on $A$ can be extended to a smooth function on $M$. We want to show that $A$ is closed. How might we proceed with a proof?
Since $M$ is a manifold, $A$ is Hausdorff. Thus if $A$ is closed if it is compact. To construct a finite cover of $A$ by open sets, let $f:A\to\mathbb{R}^k$ be a smooth function on $A$ so $f$ extends to a smooth function $F:M\to\mathbb{R}^k$ on $M$. Consider an arbitrary cover $C=\{U_\alpha:\alpha\in X\}$ of $A$. We must show that $C$ has a finite subcover.
I am not sure now how to proceed. I would like to think there is a way to use the fact that for all open sets $B\subseteq M$ such that $A\cap B$ is nonempty, we can let $F|_B$ be the restriction of $F$ to $B$ and get a finite cover of $A$ by pulling back the image of the intersection. (This isn't sufficient, it's just a sense that we should look at images in $\mathbb{R}^k$.)
What are some ideas that I can use to get started?
Best Answer
The statement in the title of this question is wrong. Consider $A=M=\mathbb{R}$. Then, clearly any smooth function on $A$ can be extended to one on $M$. But $A$ is not compact.
Of course, $A$ is still closed. Hence, this example does not disprove the statement from your post. It does, however, show that your approach (showing closedess by showing compactness) cannot succeed.
With regards to the statement in the body of the question: I would suggest trying to show closedness of $A$ more directly. Or, equivalently, openness of $M\setminus A$: I.e. take any point $x \in M\setminus A$ and try to show that there exists an open neighborhood of $x$ which does not intersect $A$.
To start you might just use an open neighborhood you get from $M$ being a manifold. This neighborhood might, of course, intersect $A$. But we also haven't used the assumption that we can extend smooth functions on $A$ yet. So we have to find some smooth function such that it's extendability says something interesting.
Maybe some function that we know cannot be extended at $x$? Then, the fact that its restriction to $A$ can be extended to the whole of $M$ (including $x$) might say something about how "close" $A$ and $x$ can get.
As a meta-hint (assuming that this is an exercise from a text book or lecture): Since compactness is a stronger property than closedness (in a Hausdorff space) it is unlikely that your type of approach would be the intended one even if one does not know the above counter example (as, then, the exercise would most likely have been stated as "Show that $A$ is compact.")
Of course, this doesn't guarantee that the approach cannot work. But one should at least be skeptical about whether it can be the right direction.