I was trying to understand this proof of the following result:
If $f \in \mathcal{C}^\infty (\mathbb{R})$ satisfies that for every $x > \in \mathbb{R}$ there exists a $n \in \mathbb{N}$ such that $f^{(n)}(x)=0$, then $f$ is a polynomial.
However I don't really understand how Baire's category theorem is applied. Could anyone explain how is it applied?
Now, I was thinking that if we could state that $\operatorname{int}X \not= \emptyset$ then by Baire's category theorem, $X \cap S_n \not= \emptyset$ for some $n \in \mathbb{N}$ and then we have a contradiction because if we take $x \in \operatorname{int}(X \cap S_n)$ then there exists an interval $(a,b)$ that contains $x$ such that $$(a,b) \subset X \cap S_n$$ so $(f^{(n)})|_{(a,b)}\equiv0$ and then $f|_{(a,b)}$ is a polynomial getting a contradiction with $x \in X$.
So, does any one have any idea if it is possible to prove that $X$ has non empty interior to conclude the proof in this way?
Here's how the theorem is applied:
We make sure that $X$ is a non-empty complete metric space, and $\{X\cap S_n\}$ is its countable closed cover. Here, $closed$ is to be understood with respect to $X$. From Baire's theorem, we deduce that some $X\cap S_n$ contains a non-empty open set. Here, $open$ is to be understood with respect to $X$. Open sets in $X$ are of the form $U\cap X$, where $U$ is open in $\mathbb{R}$. So we deduce that some open interval $(a, b)$ has a non-empty intersection with $X$, and $$(a, b)\cap X\subset X\cap S_n.$$