TITLE: How to show that the chromatic number > aleph_0
QUESTION [7 upvotes]: 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?

REPLY [6 votes]: There is a beautiful argument going back to, I think, Galvin. 
Assume that $F:V\to\omega$ is a good coloring. By transfinite recursion define $f_\alpha:\alpha\to\omega$ as follows. $f_0=\emptyset$. If $\alpha$ is limit, set $f_\alpha=\bigcup\{f_\beta:\beta<\alpha\}$. If $f_\alpha$ is given, let $f_{\alpha+1}$ be that extension of it to $\alpha+1$ for which $f_{\alpha+1}(\alpha)=F(f_\alpha)$ holds. Now one can show that each $f_\alpha$ is in $V$ and $f_{\beta}\subseteq f_\alpha$ for $\beta<\alpha$. But then $\bigcup\{f_\alpha:\alpha<\omega_1\}$ would be an injection $\omega_1\to\omega$, an impossibility.