Are $[0,1]\times[0,1]$ and $[0,1]$ order isomorphic?

Question

The ordering on the first set is reverse lexicographic and the order of the second set is the usual ordering of reals.

My intuition tells me that the answer is no. The only isomorphism between those two sets that I know of is the one that interleaves the decimal digits of the numbers in a pair (the elements of the first set) and that one definitely doesn't preserve the order.

To do this I would have to attach the decimal digits of the first number to the end of the decimal digits of the second one, and that can't really be done since there are infinitely many of them.

This is far from formal, so I would appreciate some help.

Answer 1

The answer is indeed negative. A relatively elementary solution would be to note that $[0,1]\times[0,1]$ can be partitioned into uncountably many disjoint non-degenerate intervals (they contain at least two points each).

If there was an order isomorphism, then the partition would induce a partition of $[0,1]$ into uncountably many intervals (note that the order isomorphism maps intervals to intervals).

However in every interval in $[0,1]$ there is a rational number, so there is no uncountable partition into uncountably many disjoint intervals.

(Not to mention that those intervals would be closed, and if $[0,1]$ is partitioned into closed intervals, then the Baire category theorem tells us that it is necessarily a one-part partition, namely $\{[0,1]\}$.)

Answer 2

An easy way to see that they are not isomorphic is to note that in $[0,1]\times [0,1]$ the subset $\{(1/2,x):x\in [0,1]\}$ is uncountable and discrete in the order topology..