Baire: Show that $f\colon \mathbb{R}\to\mathbb{R}$ is a polynomial in an open bounded set 
Let $f\colon\mathbb{R}\to\mathbb{R}$ be an infinitely differentiable function, and suppose that for each $x\in\mathbb{R},$ $\exists\  n=n(x)\in\mathbb{N}$ such that $f^{(n)}(x)=0$. For each fixed $n\in\mathbb{N}$, consider the sets $F_n=\{x\in\mathbb{R} : f^{(n)}(x)=0\}$ and let $I=[a,b]$ a finite closed interval.

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*Show that $$I=\bigcup_{n=0}^\infty(I\cap F_n).$$

*Show that there is a non empty open set $A\subset I$ where $f$ is a polynomial.


My attempt:

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*If $x\in I$ then by hypotesis there is $n$ such that $f^{(n)}(x)=0$ then $x\in F_n$ so then $\bigcup F_n\supset I$ then $$I=I\cap\bigcup_{n=0}^\infty F_n=\bigcup_{n=0}^\infty(I\cap F_n).$$


*Let $$B_k=\bigcap_{n=k}^\infty F_n.$$
I want to show that $I\cap B_k$ has non-empty interior for some $k\in\mathbb{N}$. Then there is $A\subset B_k$ and by Taylor's Theorem $f$ will be a polynomial. But I didn't know how to prove this. I tried using the following bigger set:
$$B=I\cap\bigcup_{k=0}^\infty B_k=I\cap\bigcup_{k=0}^\infty \bigcap_{n=k}^\infty F_n.$$
I think that it suffices to show that $B^o\neq\emptyset$ maybe using the Baire's Theorem proving that $B$ is a not meagre set.
More general Problem https://mathoverflow.net/questions/34059/
Let $f$ be an infinitely differentiable function on $[0,1]$ and suppose that for each $x \in [0,1]$ there is an integer $n \in \mathbb{N}$ such that $f^{(n)}(x)=0$. Then does $f$ coincide on $[0,1]$ with some polynomial? If yes then how.
I thought of using Weierstrass approximation theorem, but couldn't succeed.
Answer:
The proof is by contradiction. Assume $f$ is not a polynomial.
Consider the following closed sets: $$ S_n = \{x: f^{(n)}(x) = 0\} $$ and $$ X = \{x: \forall (a,b)\ni x: f\restriction_{(a,b)}\text{ is not a polynomial} \}. $$
It is clear that $X$ is a non-empty closed set without isolated points. Applying Baire category theorem to the covering $\{X\cap S_n\}$ of $X$ we get that there exists an interval $(a,b)$ such that $(a,b)\cap X$ is non-empty and $$ (a,b)\cap X\subset S_n $$ for some $n$. Since every $x\in (a,b)\cap X$ is an accumulation point we also have that $x\in S_m$ for all $m\ge n$ and $x\in (a,b)\cap X$.
Now consider any maximal interval $(c,e)\subset ((a,b)-X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$. Therefore $f^{(d)}=\mathrm{const}\neq 0$ on $[c,e]$. Hence $d< n$. (Since either $c$ or $e$ is in $X$.)
So we get that $f^{(n)}=0$ on $(a,b)$ which is in contradiction with $(a,b)\cap X$ being non-empty.
 A: Your solution for 1. is perfectly fine, the one for 2. doesn't seem to lead too far and seems a bit too complicated. You're right, Baire's theorem is the main idea:
You know that $I = \bigcup_{n=0}^\infty I \cap F_n$. Since $f^{(n)}$ is continuous, $F_n$ is closed, so $I$ is a countable union of closed sets. By Baire's theorem there must be $N$ such that $F_N$ has non-empty interior, for otherwise $I$ would be meager.
Let $J$ be a non-trivial open interval contained in $F_N$. By definition $f^{(N)}(x) = 0$ for all $x \in J \subset F_N$. On an interval a function with the property that the $N$th derivative vanishes must be a polynomial of degree at most $N-1$, so $f$ is a polynomial on $J$.
Notice that $f^{(N)} \equiv 0$ on $J$ implies that $f^{(n)} \equiv 0$ on $J$ for all $n \geq N$ because $J$ is open, so there is no need to consider your sets  $B_k$.
In fact, with some more work F. Sunyer i Balaguer and E. Corominas showed that $f$ must be a polynomial on all of $I$, see this question on mathoverflow for outlines and references.
