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I beg your pardon for poor editing, I'm just Latex beginner. This exercise comes from book on set theory.

Prove that for any finite set A with any partial order $r$, there exists $X\subset N \setminus \{0\}$ with divisibility relation | , such that sets $\langle A ,r\rangle$ and $\langle X ,|\rangle$ are isomorphic.

The answer book suggests defining mapping $a \mapsto p(a)$ and then a function :

$f\colon A \rightarrow X$

$f(a) = \prod_{a\leq x}\ p(x)$

It suffices to show that this function establishes an isomorphism in between set $A$ and that is clear.

Question:

I'd like to ask why can't we use more trivial mapping. My first idea was to use the following formula: $a \mapsto 2^a$ and then define a function based on the product in the same way they did in the book. As far I can see set of powers of $2$ does the same job as the set of primes. On the other hand I suppose similiar answer would show up in the answer key if it were correct. Could someone tell me what I'm doing wrong? Is it because we are not told that $A$ consists of integers?

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  • $\begingroup$ If you click edit, you'll be able to see some of the changes I made to your typesetting. Hopefully you can see how it looks a bit more professional now. $\endgroup$ – Dan Rust Feb 4 '15 at 18:13
  • $\begingroup$ Thanks, it's much better now. Couple more days and I'll get better :) $\endgroup$ – Pasato Feb 4 '15 at 18:17
  • $\begingroup$ yeah it's easy to get to grips with after you've used it a few times. You tend to learn little tricks along the way, like how \colon has better spacing than if you just use : Such as $f\colon X\to Y$ is better spaces than $f:X\to Y$. $\endgroup$ – Dan Rust Feb 4 '15 at 18:19
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Of course the $a's$ need not be integers, but even if you mapped $a$s to integers, say $a\mapsto g(a)$, with the mapping $a\mapsto 2^{g(a)}$, the corresponding divisibility order is a total order, whereas a general partial order need not be totally ordered.

| cite | improve this answer | |
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  • $\begingroup$ Thanks. That would mean that my mapping makes every two elements of A comparable whereas it doesn't have to be a case? $\endgroup$ – Pasato Feb 4 '15 at 18:14
  • $\begingroup$ Or in other words : it would be correct if we were told there was a total order on A ? $\endgroup$ – Pasato Feb 4 '15 at 18:20
  • $\begingroup$ Yes, I think it could work if there were a total order on $A$. $\endgroup$ – paw88789 Feb 4 '15 at 18:24

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