3 fix TeX

The proof is by contradiction. Assume $f$ is not a polynomial.

Consider the following closed sets: $$S_n = \{x: f^{(n)}(x) = 0\}. 0\}$$ and $$X = \{x: \forall (a,b)\ni x:\; f|_{(a,b)}\; is\; not\; x: f\restriction_{(a,b)}\text{ is not a \;polynomial polynomial} \}.$$

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 $f^{(d)}=const\neq f^{(d)}=\mathrm{const}\neq 0$ 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.

2 added 101 characters in body

The proof is by contradiction. Assume $f$ is not a polynomial.

Consider the following closed sets $$S_n = \{x: f^{(n)}(x) = 0\}.$$ and $$X = \{x: \forall (a,b)\ni x:\; f|_{(a,b)}\; is\; not\; a \;polynomial \}.$$

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 $x$. x\in (a,b)\cap X$. Now consider and any maximal interval$(c,e)\subset ((a,b)-X)$. Recall that$f$is a polynomial of some degree$d$on$(c,e)$. Therefore$f^{(d)}\neq f^{(d)}=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)$which contradicts the definition of is in contradiction with$X$.(a,b)\cap X$ being non-empty.

1

The proof is by contradiction. Assume $f$ is not a polynomial.

Consider the following sets $$S_n = \{x: f^{(n)}(x) = 0\}.$$ $$X = \{x: \forall (a,b)\ni x:\; f|_{(a,b)}\; is\; not\; a \;polynomial \}.$$

It is clear that $X$ is a non-empty closed set without isolated points. Applying Baire category theorem to $X$ we get that there exists $(a,b)$ 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 have that $x\in S_m$ for all $m\ge n$ and $x$.

Now consider and any maximal interval $(c,e)\subset ((a,b)-X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$. Therefore $f^{(d)}\neq 0$ 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 contradicts the definition of $X$.