Projective limit of system sets compatible maps $\phi : \{x_1, \dots, x_n \} \hookrightarrow \mathbb{Z}$

Question

This question is about the projective limit of the following system:

• Let $I$ be the poset of finite subsets of $\mathbb{R}$ partially ordered by inclusion.
• Let $S_U$ be the set of injective maps $\phi : U \hookrightarrow \mathbb{Z}$.
• For $U \subseteq V$, let $f_{VU} \ : \ S_V \longrightarrow S_U \ : \ \phi \longmapsto \phi|_U$

I have to show that the projective limit $\varprojlim_{W \in I} S_W \ = \ \emptyset$. How can I do this?

---

My own thoughts

It is easy to show that the definitions above give rise to a projective system. According to the notes I work with the projective limit is of the system above is: $$ \left\{ (\phi_W)_{W \in I} : \forall (U, V \text{ s.t. } U \subseteq V), \ f_{VU}(\phi_V) = \phi_U \right\} $$

Let's look at $(\phi_W)_{W\in I} \in \varprojlim_{W \in I} S_W $. For any finite $U,V \subseteq \mathbb{R}$ the maps $\phi_U$ and $\phi_V$ have to agree on $U \cap V$. This makes me think that the $(\phi_W)_W$ corresponds with a map $\phi : \mathbb{R} \rightarrow \mathbb{Z}$. As a matter of fact, thanks to the compatibility constraints we could just define: $$ \phi \ : \ \mathbb{R} \longrightarrow \mathbb{Z} \ : \ x \longmapsto \phi_{\{x\}}(x) $$ This makes me think that the projective limit we consider is the set of injective maps $\phi : \mathbb{R} \rightarrow \mathbb{Z}$. Since $\mathbb{R}$ isn't countable such maps don't exist.

---

What does actually justify the last thought? I thought of proving the universal property of $\{ \phi : \mathbb{R} \hookrightarrow \mathbb{Z} \}$ with maps $$ f_i \ : \ \{ \phi : \mathbb{R} \hookrightarrow \mathbb{Z} \} \ \longrightarrow \ \varprojlim_{W \in I} S_W \ : \ \phi \ \longmapsto \ \phi|_W $$ But I failed.

Answer 1

You have the right idea: given $(\phi_W)_{W\in I}\in\varprojlim_{W\in I} S_W$, you have described how to define a function $\phi:\mathbb{R}\to\mathbb{Z}$ by the formula $\phi(x)=\phi_{\{x\}}(x)$. Now all you have to show is that $\phi$ is injective to get a contradiction. The key point is that failure of injectivity can be detected by a finite set. That is, if $\phi$ were not injective, there would be $x,y\in\mathbb{R}$ with $x\neq y$ and $\phi(x)=\phi(y)$. Now just consider $\phi_{\{x,y\}}$: we must have $$\phi_{\{x,y\}}(x)=\phi_{\{x\}}(x)=\phi(x)=\phi(y)=\phi_{\{y\}}(y)=\phi_{\{x,y\}}(y).$$ This contradicts the fact that $\phi_{\{x,y\}}$ must be injective.