$\mathbb R\times\mathbb R$ not order isomorphic to $\mathbb R$

Question

I am trying to show that if you use the lexicographic ordering induced by $\mathbb R$ on $\mathbb R\times\mathbb R$ they are not isomorphic.

Is it enough to use the counter-example that whilst $(1,0) < (1,1)$ in $\mathbb R\times \mathbb R$, $1$ is not strictly less than $1$ in $\mathbb R$?

Thanks

Answer 1

Hint: the fundamental property of the order on the reals is that every subset that has an upper bound has a lowest upper bound.

Is this property shared by the product under the lexicographic order?

Your example is not a proof: it's obvious that an order preserving bijection must have strange properties, if it existed.

Answer 2

No: that merely shows that certain attempts to build an isomorphism can’t work. You need to show that no map from $\Bbb R\times\Bbb R$ to $\Bbb R$, no matter how weird, is an order-isomorphism.

HINT: For $x\in\Bbb R$ let $I_x=\{x\}\times(0,1)$; this is an open interval in $\Bbb R\times\Bbb R$. $\{I_x:x\in\Bbb R\}$ is therefore an uncountable family of pairwise disjoint open intervals in $\Bbb R\times\Bbb R$. Does such a family exist in $\Bbb R$? Remember, ever non-empty open interval in $\Bbb R$ contains a rational.