Let $V =\{ f | \exists \_{\alpha<\omega\_1} ( f:\alpha \rightarrow \mathbb{N} \wedge f $ is $ 1-1) \}$. We define $E\subseteq [V]^2 $, such that $\forall\_{f,g\in V } (<f,g>\in E \longleftrightarrow ( f\subseteq g \vee g\subseteq f))$. We need to show that the chromatic number of $G=(V,E)$ is bigger than $\aleph\_ 0$. It is obvious why it cannot be $<\aleph\_0$, but I'm having trouble showing that it cant be $=\aleph\_0$. Could anyone help?



How to show that the chromatic number > aleph_0
Consider any coloring $\varphi:V\to\omega$. Construct a strictly decreasing sequence $\langle X\_n:n\lt\omega\rangle$ in $\mathcal P(\mathbb N)$ and a strictly increasing sequence $\langle f\_n:n\lt\omega\rangle$ in $V$ so that for each $n\in\omega$ we have: 


* $\text{range}(f\_n)\subset X\_n$,
* $|X\_n\setminus\text{range}(f\_n)|=\aleph\_0$,
* either $\varphi(f\_n)=n$, or else $\varphi(f)\ne n$ for all $f\in V$ such that $f\_n\subseteq f,\text{range}(f)\subset X\_n$, and $|X\_n\setminus\text{range}(f)|=\aleph\_0$.


[The last condition means that, whenever we can, we take $f\_n$ to be a vertex of color $n$.]


Let $f=\bigcup\_{n\lt\omega}f\_n\in V$ and let $n=\varphi(f)$. Then $\varphi(f\_n)=n=\varphi(f)$, showing that $\varphi$ is not a proper coloring of $G$. 


[Note that $\text{range}(f)\subseteq X=\bigcap\_{n\lt\omega}X\_n$, and $X\_n\setminus X$ is infinite for every $n$ as the $X\_n$ are strictly decreasing.]


**P.S.** In other words, the partially ordered set $P=(V,\subseteq)$ has the partition property $P\rightarrow(2)\_{\omega}^1$.  

A similar argument shows that $P\rightarrow(\omega)\_{\omega}^1$, which in turn implies (according to something I read somewhere) that $P\rightarrow(\alpha)\_{\omega}^1$ 
for every ordinal $\alpha\lt\omega\_1$. [Posets $P$ such that $P\rightarrow(\omega)\_{\omega}^1$ are called *non-special*, see [this answer](https://mathoverflow.net/questions/138651/a-problem-about-ramsey-property/138659#138659).]


**P.P.S.** To prove $P\rightarrow(\omega+1)\_{\omega}^1$ we proceed as before except that, instead of trying to choose $f\_n$ of color $n$, we try to choose $f\_n$ of color $c\_n$ where $\langle c\_n:n\lt\omega\rangle$ is an enumeration of $\omega$ with each element repeated infinitely often.


