Smooth functions can not be non-polynomial at one point I am studying the classic problem here about Baire's Category theorem. 
One of the remarks is that if $f$ is smooth and not a polynomial,
$$X = \big\{x : \forall(a,b)\ni x, f|_{(a,b)} \text{ is not a polynomial}\big \}$$
has no isolated points. How could I show that the smooth function $f$ can not be non-polynomial at one (or isolated) point.
Intuitively a function $f$ can be non-polynomial at a single isolated point $x$ if there is a corner in some derivative $f^{(n)}$ at $x$. And when the function is smooth, this can not happen. 
 A: If false, then there is $\delta > 0$ such that $(x,x+\delta]\cap X = \emptyset.$ Because $x + \delta \not \in X,$ there is an open interval $I$ centered at  $x + \delta$ such that $f$ equals a polynomial $p$ in $I.$ We can assume the left end point of $I$ is greater than $x.$ Then there is an open interval $J$ centered at this endpoint such that $f$ equals a polynomial $q$ in $J.$ This forces $p=q$ everywhere. Why? Because they both equal $f$ in $I\cap J,$ which is open and non-empty, and two polynomials that agree on an infinite set are equal everywhere. It follows that there is a polynomial that equals $f$ on $I\cup J.$
This process can be continued, obtaining larger open intervals on which $f$ is a polynomial. In fact this will lead to one polynomial that equals $f$ on $(x,x+\delta]$ (that needs proving, but I'll leave it for now). This implies there exists $m\in \mathbb N$ such that $f^{(m)} \equiv 0$ on $(x,x+\delta].$ By continuity, $f^{(m)}(x) = 0$ as well.
The same argument applies to the left of $x,$ where we conclude there is $\epsilon >0$ and $n\in \mathbb N$ such that $f^{(n)} \equiv 0$ on $[x-\epsilon,x].$ Let $N = \max (m,n).$ Then $f^{(N)} \equiv 0$ on an open interval containing $x,$ and therefore $f$ is a polynomial of degree $< N$ on this interval. That's a contradiction, giving the result.
