Prove there's no partition of $\omega_1\times \omega_1$ with the following property

Question

We define $\prec$ on $\omega_1\times\omega_1$ by $(a,b)\preceq(c,d)\iff a\leq c\land b\leq d$.

Then $x\prec y\iff x\preceq y\land x\neq y$.

Now prove there's no partition of $\omega_1\times\omega_1$ of countably many $X_i$'s where $(X_i,\prec)$ is total for every $i$.

I was thinking of finding an uncountable subset $A$ of $\omega_1\times\omega_1$, then we know there is $x,y\in A$ where $x,y\in X_i$ for some $i$. Then we prove that $X_i$ is not total. But I can't find such a set $A$.

Answer 1

HINT: Suppose that $\{X_k:k\in\omega\}$ is a partition of $\omega_1\times\omega_1$. For each $\alpha\in\omega_1$ there is a $k(\alpha)\in\omega$ such that $|\{\beta\in\omega_1:\langle\alpha,\beta\rangle\in X_{k(\alpha)}\}|=\omega_1$. Choose distinct $\alpha_0,\alpha_1\in\omega_1$ such that $k(\alpha_0)=k(\alpha_1)=k$, say, and show that there are $\beta_0,\beta_1\in\omega_1$ such that $\langle\alpha_i,\beta_i\rangle\in X_k$ for $i\in 2$, $\alpha_0<\alpha_1$, and $\beta_0>\beta_1$. Conclude that $\preceq$ does not linearly order $X_k$.

Answer 2

Correction: I wrote this thinking that $x\prec y\iff x_1<y_1\land x_2<y_2$, but that is only a special case of the ordering relation defined in the original post. Generalizing the proof more or less follows the reasoning in Brian M. Scott's answer, so I will leave it as is.

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Let $X=\bigcup_{i\in I}\{X_i\}$ be a countable partitioning of $\omega_1\times\omega_1$.

Select an element $x\in\omega_1$ and let $A=\bigcup_{i\in I}\{A_i\}$ be a countable partitioning of $\{x\}\times (\omega_1\setminus\{x\})$ such that $A_i\subset X_i$ for all $i\in I$.

Every subset of a totally ordered set is totally ordered. If the order on $X_i$ is total, it follows that the order on $A_i$ is also total, whence $u\prec v$, $v\prec u$, or $u=v$ for any $u,v\in A_i$.

Let $f:\omega_1\setminus\{x\}\to A$ be the function defined by the expression:

$$f(a_i)=j\iff (x,a_i)\in A_j$$

Notice that $f$ is never a bijection (this follows from the fact that $A$ is countable) - hence there must exist some triple $i,j,k$ where $i\ne j\ne k$ such that $f(a_i)=f(a_j)=k$, whereby $(x,a_i),(x,a_j)\in A_k$.

Because $(x,a_i)\nprec(x,a_j)$, $(x,a_j)\nprec(x,a_i)$, and $(x,a_i)\ne(x,a_j)$, the order on $A_k$ cannot be total. Because $A_i\subset X_i$ for all $i\in I$, the order on $X_i$ is not total (contrapositive).