For every poset $(X,\le)$ there is a linear order $\preceq$ on $X$ which extends $\le$

Question

Use the Compactness Theorem in order to show that for every partially ordered set $(X,\le)$ there is a linear order $\preceq$ on $X$ which extends $\le$, that is: for all $x,y\in X$ we have $x\le y \to x\preceq y$.

My attempt:

Useful fact. Every set $X$ admits a linear order.

Consider the language $L = (\le, \preceq, \{c_x\mid x\in X\})$, where $\le$ is given, $\preceq$ is the linear order that can be found for $X$ and $c_x$ constants. Define the $L$-theory $$T = \{ \text{axioms of linear order for} \preceq\}\cup \{\forall x,y\in X: x\le y\to x\preceq y\}\cup \{\neg(c_x=c_y)\mid x,y\in X, x\ne y\}.$$ Our goal is to show that $T$ is consistent. Let $T'\subseteq T$ be a finite subset of $T$. Goal: $T'$ is consistent.

We have $T'\subseteq \{\text{axioms}\}\cup \{\forall x,y\in X': x\le y\to x\preceq y\}\cup \{\neg(c_x=c_y)\mid x,y\in X', x\ne y\}$ for some finite $X'\subseteq X$. Then, $X'$ admits a linear order $<_{X'}$ (by the fact), and in particular $X'$ is partially ordered by $\le$. Can I choose/define $\preceq := <_{X'}$. In that case, $M=X'$ with $<^M = <_{X'}$ and $c_x^M=x$ if $x\in X'$, $c_x^M=x_0\in X' $ arbitrary if $x\in X-X'$, is a model for $T'$. We conclude that $T$ has a model $N$ as well. The map $X\to N: x\mapsto c_x^N$ is injective. We know $\forall n,n'\in N: n\le n'\to n\preceq n'$. I believe that the injective map transfers this property to $X$ as well.

Is this alright?

Thanks.

Answer 1

When building the model for the finite theory $T'$, you do not prove that $x\leq y \rightarrow x\preceq y$ for every $x,y\in X'$. To do this, you need $\preceq$ to extend $\leq$ on $X'$, rather than be arbitrary.

You need the following lemma :

Every finite partial order can be extended to a linear order.

The rest of the proof is ok.