One way to phrase van der Waerden's Theorem is:


*For every finite coloring of $\mathbb N$ and every finite $F \subseteq \mathbb N$, there exist $a,b \in \mathbb N$ such that $a + b \cdot F$ is monochromatic.*


My question is whether the same thing is true for the first uncountable ordinal $\omega\_1$. Specifically, is it true that:


**$(\*)$** *For every finite coloring of $\omega\_1$ and every finite $F \subseteq \omega\_1$, there exist $\alpha,\beta \in \omega\_1$ such that $\alpha + \beta \cdot F$ is monochromatic.*


The operations $+$ and $\cdot$ are the usual operations of [ordinal arithmetic](http://en.wikipedia.org/wiki/Ordinal_arithmetic).




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*Remark 1:* There is more than one way to generalize the statement of van der Waerden's Theorem to $\omega\_1$, but this one seems the most interesting from the point of view of combinatorics. If you restrict $F$ to being a finite interval, then $(\*)$ is simply true by the usual van der Waerden Theorem (and we can always take $\alpha, \beta \in \omega$). If we let $F$ range over all countable subsets of $\omega\_1$, then $(\*)$ cannot be proved from ZFC. This doesn't make it uninteresting, and in fact I've asked another question about that version [here](https://mathoverflow.net/questions/206935/is-this-version-of-van-der-waerdens-theorem-consistent-with-zfc). I also don't know what happens when we use countable colorings instead of finite colorings.


*Remark 2:* This question arose while I was wondering whether the ultrafilters proof of van der Waerden's Theorem generalizes nicely to other settings. That same kind of proof doesn't seem to give us anything about $(\*)$, but I also can't think of any obvious counterexamples.



Does van der Waerden's Theorem hold for $\omega_1$?
The answer is no; this generalization is inconsistent, even with just two colors,
and with $F=\{\omega,\omega^2\}$ of size two.


**Theorem.** There is a coloring of ordinals with two colors, such that for any ordinals $\alpha$ and $\beta$, the ordinals $\alpha+\beta\cdot\omega$ and $\alpha+\beta\cdot\omega^2$ get different colors. 


**Proof.** Given an ordinal $\eta$, look at its [Cantor normal
form](http://en.wikipedia.org/wiki/Ordinal_arithmetic#Cantor_normal_form) $$\eta=\omega^{\eta\_n}+\cdots+\omega^{\eta\_0}\qquad\text{
with }\eta\_n\geq\cdots\geq \eta\_0.$$ Let us color $\eta$ with the
even/odd parity of the last exponent $\eta\_0$.


Suppose that $\alpha$ and $\beta$
are ordinals. I claim that $\alpha+\beta\cdot\omega$ and
$\alpha+\beta\cdot\omega^2$ get different colors.


To see this, let $\beta=\omega^{\beta\_n}+\cdots+\omega^{\beta\_0}$
be the Cantor normal form of $\beta$, with $\beta\_n\geq\cdots\geq\beta\_0$. Note that
$\beta\cdot\omega=\omega^{\beta\_n+1}$, because $\omega$ copies of
$\beta$ is at least this big, and conversely also
$$\beta\cdot\omega\leq(\omega^{\beta\_n}\cdot
(n+1))\cdot\omega=\omega^{\beta\_n}\cdot((n+1)\cdot\omega)=\omega^{\beta\_n}\cdot\omega=\omega^{\beta\_n+1};$$
so they are equal. It follows that 
$\beta\cdot\omega^2=\omega^{\beta\_n+2}$.


Thus, 
$\alpha+\beta\cdot\omega=\alpha+\omega^{\beta\_n+1}$, and
furthermore, $\omega^{\beta\_n+1}$ will be the final term in the
Cantor normal form of this ordinal, because the smaller terms of
$\alpha$, if any, will simply be absorbed into it.


Similarly, $\alpha+\beta\cdot\omega^2=\alpha+\omega^{\beta\_n+2}$,
and $\omega^{\beta\_n+2}$ will be the final term in the Cantor
normal form of this ordinal also.


So the proof is completed by noting that $\beta\_n+1$ and
$\beta\_n+2$ have opposite parities, and so
$\alpha+\beta\cdot\omega$ and $\alpha+\beta\cdot\omega^2$ get
different colors under the coloring I defined. **QED**


**Update.** Here is another coloring that works with $F=\{\omega,\omega+\omega\}$. 


**Theorem.** There is a coloring of ordinals with two colors such that for any ordinals $\alpha$ and $\beta$, the ordinals $\alpha+\beta\cdot\omega$ and $\alpha+\beta\cdot(\omega+\omega)$ get different colors. 


**Proof.** Color any ordinal $\eta$ by the parity of the number of terms in the Cantor normal form $\eta=\omega^{\eta\_n}+\cdots+\omega^{\eta\_0}$, where $\eta\_n\geq\cdots\geq\eta\_0$. Suppose that $\alpha$ and $\beta$ are any two ordinals. As noted above, $\beta\cdot\omega$ has only one term in the Cantor normal form, and so $\beta\cdot(\omega+\omega)=\beta\cdot\omega+\beta\cdot\omega$ has precisely two (identical) terms. It follows that the Cantor normal forms of the ordinals $\alpha+\beta\cdot\omega$ and $\alpha+\beta\cdot(\omega+\omega)$ differ by exactly one term, and so they get different colors. (Note that the terms of $\alpha$ that are smaller than $\beta\cdot\omega$ cancel the same in both cases, and the larger terms don't cancel.) **QED**


