Dictionary ordering in $\mathbb{R}^2$ is not complete.

Question

Show that the order "$\leq$" on $\Bbb{R}^2$ defined by,
$(a,b)\leq(c,d)$ if ($a<c$) or, $(a=c$ and $b\leq d)$ is not complete.

Hint: Use the set $E=\{(\frac1 n, 1-\frac1 n): n\in \Bbb{N}\}$.

Can any one help me with this? How can the set $E$ be used to show that, the ordering on $\mathbb{R}^2$ is not complete? Thank you.

Answer 1

The set should be

$$E=\big\{(1-1/n, 1/n)\ \big|\ n\in\mathbb{N}\big\}$$

I've swapped coordinates (your original $E$ has $(1,0)$ as the least upper bound).

With that assume that $(p,q)$ is an upper bound of $E$. By looking at the first coordinate we get that $1-1/n\leq p$ for any $n\in\mathbb{N}$. Therefore we conclude that $p\geq 1$. But then we have

$$(1-1/n,1/n)<(1,r)$$

for any $n\in\mathbb{N}$ and any $r\in\mathbb{R}$ by our order definition. In particular if $(p,q)$ is an upper bound of $E$ then $(1,q-1)$ is an upper bound of $E$ which is lower than $(p,q)$. Therefore the least upper bound of $E$ does not exist.