2
$\begingroup$

I found that the usual ordinal exponentiation $\alpha^{\beta}$ is the set of functions from $\beta$ to $\alpha$ with finite support, ordered by antilexicographic order. (least significant position first) It cannot be defined on whole functions, since $f,g:\omega \to 2$ given by $f=(0,1,0,1,\cdots)$ and $g=(1,0,1,0,\cdots)$ are not comparable.

Then how about the lexicographic order? Isn't it possible to define well order on the set of all functions from $\beta$ to $\alpha$?

$\endgroup$

1 Answer 1

8
$\begingroup$

The lexiographic order defines a linear order on ${}^\beta\alpha$, but in general, it is not a well-ordering. For example, in the case ${}^\omega2$, the set $$ \{ (1,0,\ldots), (0,1,0, \ldots), (0,0,1,0,\ldots), \ldots \} $$ does not have a lexicographic least element.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .