0
$\begingroup$

I would like to ask how many well-ordering of set $\mathbb N$ exist. And they shouldn't be isomorphic to each other. I found out that the set of well ordering of set $A$ is the subset of $A \times A$. And that the set can be always well-ordered, because it is enough to do an injection between these set and ordinal numbers. But how should I find out how many well-ordering of the set $\mathbb N$ exist? :))

$\endgroup$
8
  • $\begingroup$ Try to give your question a title that is descriptive. This will not only help you find other, similar questions which may already have an answer, but will also help other users know what your question is about before actually reading it. $\endgroup$
    – Asaf Karagila
    Jun 3, 2020 at 10:39
  • $\begingroup$ Hello. Ok, thank you. I didn't know that the topis is the same but it is good for me too, because now I know much more answers :) Hi :) $\endgroup$
    – lida
    Jun 3, 2020 at 10:55
  • $\begingroup$ And can I ask, here was the answer but it isn't here now. Would you know where can I find it? :) $\endgroup$
    – lida
    Jun 3, 2020 at 11:14
  • $\begingroup$ Two answers on this page were deleted by their authors. Perhaps because they felt they did not address the actual question, or perhaps they felt that the duplicate should sufficiently answer your query. $\endgroup$
    – Asaf Karagila
    Jun 3, 2020 at 11:15
  • $\begingroup$ I also don't know why you've edited your question to add a few empty lines at the bottom, but this is not something we expect people to do. Please don't do that in the future. $\endgroup$
    – Asaf Karagila
    Jun 3, 2020 at 11:16

2 Answers 2

3
$\begingroup$

First, $\mathbb{N}$ comes with a natural well-order, which is the usual $\leq$.

Let $(\mathbb{N}, R)$ be a well-ordered set. Then we have an associated ordinal $O(\mathbb{N}, R) \cong (\mathbb{N},R)$. Thus the associated ordinals must be countable.

Conversely, given a countable ordinal, we can lift an ordinal structure on $\mathbb{N}$.

Thus, the amount of (non-isomorphic) well-orders on $\mathbb{N}$ is the amount of countable infinite ordinals, which is $\aleph_1$ (the cardinality of the first uncountable ordinal).

$\endgroup$
26
  • $\begingroup$ Thank you for your answer. It helped me. :) $\endgroup$
    – lida
    Jun 3, 2020 at 11:25
  • $\begingroup$ You might consider accepting it then :) $\endgroup$ Jun 3, 2020 at 11:27
  • $\begingroup$ How do you thing? $\endgroup$
    – lida
    Jun 3, 2020 at 11:28
  • $\begingroup$ Under the amount of upvotes (the 2 left next to my answer) there should be a button. $\endgroup$ Jun 3, 2020 at 11:28
  • 1
    $\begingroup$ I think that now I am littlebit prepared to a test. I am really grateful for your help :) $\endgroup$
    – lida
    Jun 3, 2020 at 12:23
0
$\begingroup$

Here's a starting point:

Take the well-ordering $\preceq$ on $\mathbb N$ defined by

$a\preceq b \text{ if } \begin{cases}\text{$a$ is even and $b$ is odd}\\a\leq b \text{ where $a$ and $b$ both odd}\\ a\leq b \text{ where $a$ and $b$ both even}\end{cases}$

In effect this takes the standard well-ordering on the even natural numbers and the standard ordering on the odd natural numbers, and puts one before the other.

This is a well-ordering but is not order-isomorphic to the standard ordering $\leq$ since the number $1$ has infinitely many values less than it.

$\endgroup$
1
  • $\begingroup$ Thank you, you answered to my another question, find two well-ordering of the set $\mathbb N$ which aren't isomorphic and don't have got the largest element :) $\endgroup$
    – lida
    Jun 3, 2020 at 11:09

Not the answer you're looking for? Browse other questions tagged or ask your own question.