Is necessarily a total order on $R^n$ a bijection on $R$?

Question

I don't think it's true but: is necessarily a total order on $R^n$ equivalent to a bijection on $R$? If someone knows how to show that is true, or false...

Thanks

Answer 1

No, a total order on $\Bbb R^n$ doesn't automatically get you a bijection (in the sense of an order isomorphism) with $\Bbb R$, the way a bijection with $\Bbb R$ induces a total order.

For instance, the order on $\Bbb R^n$ could have a minimal or maximal element, or it could be incomplete or have gaps.

Answer 2

If you mean that any total order on $\mathbb R^n$ is order isomorphic to $\mathbb R$, then no. For example, in the dictionary order on $\mathbb R^2$, the $x$ axis $\mathbb R\times\{0\}$ is an uncountable discrete subset, and the $y$ axis $\{0\}\times\mathbb R$ is a bounded subset without a supremum. Neither type of subset exists in the usual order on $\mathbb R$.