Dimension of the poset $(\mathbb{R}^2,\ge)$

I am working on the question about the poset $(\mathbb{R}^2,\ge)$ such that $(x_1,y_1)\ge (x_2,y_2)$ iff $x_1\ge x_2, y_1\ge y_2$. I found the conclusion that the dimension of this poset is $2$, but I am not quite clear about how to prove it (The dimension of a poset is the smallest number of total orders the intersection of which gives rise to the partial order). Any thought would be appreciated.

Solution 1:

Great, so I think you and I are on the same page about why the dimension is at most $2$; we just need to show that the dimension can't be $1$. But if the dimension was $1$, then the partial order $(x_1, y_1) \geq (x_2, y_2)$ would have to be total, and we can check that it isn't. Indeed, $(1, -1)$ and $(-1, 1)$ are incomparable, and this is enough.

Solution 2:

As explained in Keeley's answer, the poset does not have dimension $1$, since it is not a total order. It remains to show that it has dimension $\leq 2$, and to do this, we just need to exhibit it as the intersection of two total orders.

The lexicographic order $\leq_{\mathrm{lex}}$ is defined by $(x_1,y_1)\leq_{\mathrm{lex}} (x_2,y_2)$ if and only if $x_1 < x_2$ or ($x_1 = x_2$ and $y_1 \leq y_2$).

The reverse lexicographic order $\leq_{\mathrm{rlex}}$ is defined in the same way, but reversing the roles of $x$ and $y$: $(x_1,y_1)\leq_{\mathrm{rlex}} (x_2,y_2)$ if and only if $y_1 < y_2$ or ($y_1 = y_2$ and $x_1 \leq x_2$).

If you're not familiar with these orders, you should prove that they are total. I claim that their intersection is our ordering $\leq$.

Suppose $(x_1,y_2)\leq (x_2,y_2)$. This means $x_1\leq x_2$ and $y_1\leq y_2$. Since $x_1\leq x_2$, we have $x_1 < x_2$ or $x_1 = x_2$. In the latter case, we also have $y_1\leq y_2$. So in either case, we have $(x_1,y_1)\leq_{\mathrm{lex}} (x_2,y_2)$. The symmetric argument shows that $(x_1,y_1)\leq_{\mathrm{rlex}} (x_2,y_2)$. So ${\leq} \subseteq (\leq_{\mathrm{lex}}\cap \leq_{\mathrm{rlex}})$.

Conversely, suppose $(x_1,y_1)\leq_{\mathrm{lex}} (x_2,y_2)$ and $(x_1,y_1)\leq_{\mathrm{rlex}} (x_2,y_2)$. By the definition of $\leq_{\mathrm{lex}}$, $x_1 < x_2$ or $x_1 = x_2$, so $x_1\leq x_2$. Similarly, by definition of $\leq_{\mathrm{rlex}}$, $y_1\leq y_2$. So $(\leq_{\mathrm{lex}}\cap \leq_{\mathrm{rlex}})\subseteq {\leq}$.