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In ZFC minus the power set axiom we can define $\mathbb N$, $\mathbb Z$, $\mathbb Q$ and $\mathbb Q[X]$ in the usual way. Consider the statement (*) : for $a,b\in{\mathbb Q}$ and $P\in {\mathbb Q}[X]$, if $P'\geq 0$ on $[a,b]$ (here the interval is in $\mathbb Q$ of course), then $P$ is nondecreasing on $[a,b]$.

(*) is obviously provable from ZFC, using real numbers. But is it provable from ZFC minus the power set axiom ?

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    Why couldn't you get $\mathbb{R}$ without the powerset axiom ? If you use the Cauchy sequence construction, it seems to me you never use Power Set. Maybe it is when defining $\mathbb{Q}^\mathbb{N}$ ? But then how do you define $\mathbb{Q}[X]$ ? – 2017-01-29
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    @Max. If you remove the powerset axiom, you cannot go beyond hereditarily countable sets, so you cannot reach uncountable sets. The set of Cauchy sequences is defined as a subset of ${\mathbb Q}^{\mathbb N}$, so you need this latter set. In turn, to define ${\mathbb Q}^{\mathbb N}$ you need the powerset axiom. To define ${\mathbb Q}[X]$, however, you only need ${\mathbb Q}^{< \mathbb N}$, i.e. the set of all finite sequences with values in $\mathbb Q$. – 2017-01-29
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    Oh indeed, my mistake ( in general to define $E^F$ you need powerset) – 2017-01-29

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Assume first $P'\geq\varepsilon>0$. Then for every fixed $x_0$, one can write $$ P(x)=\sum_{i=0}^ka_i^{(x_0)} (x-x_0)^i, $$ where $|a_i^{(x_0)}|\leq M$ for some $M$ uniformly in $i$ and $x_0$. Using $a_1\geq\varepsilon$, as long as $x\in[x_0,x_0+\varepsilon/kM\wedge 1]$: $$ P(x)\geq a_0+\varepsilon (x-x_0)-\sum_{i=2}^k M(\varepsilon/kM)(x-x_0)^{i-1}\geq a_0+(x-x_0)\varepsilon/k=P(x_0). $$ Now for every $y\geq x_0$ one can find $x_0\leq x_1\cdots \leq x_n= y $ with $n\leq (a-b)/(\varepsilon/kM\wedge 1)$ such that $x_i\in[x_{i+1},x_{i+1}+\varepsilon/kM\wedge 1]$, and so $P(x_{i+1})\geq P(x_i)$, which gives the claim.

If one only has $P'\geq 0$, then apply the above for $\hat P^\varepsilon(x)=P(x)+\varepsilon x$, and note that if $\hat P^\varepsilon(x)\geq \hat P^\varepsilon(y)$ for all $\varepsilon\in\mathbb{Q}$, then also $P(x)\geq P(y)$.