# Absolutely defined ordering on the set of all infinite sequences is not well ordered

From the Wikipedia page on Ordinal Exponentiation:

The definition of ordinal exponentiation for finite exponents is straightforward. If the exponent is a finite number, the power is the result of iterated multiplication. For instance, $$ω^2 = ω·ω$$ using the operation of ordinal multiplication. Note that $$ω·ω$$ can be defined using the set of functions from $$2 = \{0,1\}$$ to $$ω = \{0,1,2,...\}$$, ordered lexicographically with the least significant position first:

$$(0,0) < (1,0) < (2,0) < (3,0) < ... < (0,1) < (1,1) < (2,1) < (3,1) < ... < (0,2) < (1,2) < (2,2) < ...$$

Here for brevity, we have replaced the function $$\{(0,k), (1,m)\}$$ by the ordered pair $$(k, m)$$.

Similarly, for any finite exponent $$n$$, $$\omega^{n}$$ can be defined using the set of functions from n (the domain) to the natural numbers (the codomain). These functions can be abbreviated as $$n$$-tuples of natural numbers.

But for infinite exponents, the definition may not be obvious. A limit ordinal, such as $$ω$$, is the supremum of all smaller ordinals. It might seem natural to define $$ω$$ using the set of all infinite sequences of natural numbers. However, we find that any absolutely defined ordering on this set is not well-ordered [citation needed]. To deal with this issue we can use the variant lexicographical ordering again. We restrict the set to sequences which are nonzero for only a finite number of arguments.

There is no citation provided for the statement in bold. Can someone explain in detail why any absolutely defined ordering on the set of all infinite sequences is not well ordered? We could also add this page on Wikipedia as a citation in that case.

## 1 Answer

The set of infinite sequences of natural numbers is just the Cartesian product $$\Bbb N^{\Bbb N}$$. If $$\Bbb N$$ is given the discrete topology, this space is provably homeomorphic to the space of rational numbers as a subspace of $$\Bbb R$$. Since $$\Bbb Q$$ has a definable well-ordering, $$\Bbb R$$ would have one if $$\Bbb N^{\Bbb N}$$ did. But it is consistent with $$\sf{ZF}$$ that $$\Bbb R$$ have no well-ordering and hence that $$\Bbb N^{\Bbb N}$$ have no well-ordering. This in turn ensures that no ‘absolutely defined’ order on $$\Bbb N^{\Bbb N}$$ can be a well-ordering. The accepted answer to this question goes into more detail, and one of the set theorists can doubtless provide a better answer.