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I'm trying to solve the problem 7.3 of the book Notes on Set Theory written by Moschovakis.

Basically, I have to prove that for every poset $P$ we have $\mathrm{Succ}(P)=_o P+_o [0,1)$, were $P+_o [0,1)$ is obtained by placing disjoint copies of $P$ and $[0,1)$ side-by-side and $\mathrm{Succ}(P)$ is obtained by adding a new point to the elements of $P$.

So, to prove that I'm supposed to find an order preserving bijection $\pi$ (called similarity) between $\mathrm{Succ}(P)$ and $ P+_o [0,1)$. But it seems imposible to me to find a bijection, because if $P$ is finite, we'd have that the image of a point have to be an interval.

Am I wrong? In this case, could youu give me a bijection? Maybe I've understood wrong some definitions?

Thank you!

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    $\begingroup$ What is $[0,1)$? If that's the order set $\{x:0\le x<1\}$ within the ordinal numbers or $\Bbb Z$, then it is indeed the single point $\{0\}$. Or is that the poset's $0,1$, i.e. the greatest and least elements of the poset? Those aren't always uniquely defined though. $\endgroup$ Jan 9 '15 at 21:29
  • $\begingroup$ Not that it makes any difference to this problem, but how do you add the extra point in $\text{Succ}(P)$? Do you make it non-comparable to all elements? $\endgroup$
    – Git Gud
    Jan 9 '15 at 21:29
  • $\begingroup$ Maybe that's my problem Mario. I thought that $[a,b)=\{x\in \mathbb{R}\,:\, a \leq x < b\}$ but it's make more sense what do you say, that it's $\{0\}$. Thank you! $\endgroup$
    – MaríaCC
    Jan 9 '15 at 21:35
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To confirm what Mario Carneiro suggested in the comments, at the beginning of Chapter $7$ you’ll find that $[0,n)$ is defined to be $\{i\in\Bbb N:i<n\}$, so $[0,1)=\{0\}$. You should probably write your similarity in terms of the formal definitions of $\operatorname{Succ}(P)$ and $P+_oQ$.

$\operatorname{Succ}(P)$ is formally defined on p. $93$ as the poset whose field is $\operatorname{Field}(P)\cup\{t_P\}$, where $t_P=\mathbf{r}(\operatorname{Field}(P))$, and whose ordering is given by setting $x\le_{\operatorname{Succ}(P)}y$ iff

$$x\le_Py\lor[x\in P\land y=t_P]\lor x=y=t_P\;.$$

Thus, the new element is specifically added at the ‘top’ of $P$: it’s bigger than every element of $P$.

$P+_oQ$ is formally defined at $\mathbf{7.37}$ to be the poset whose field is

$$\big(\{0\}\times\operatorname{Field}(P)\big)\cup\big(\{1\}\times\operatorname{Field}(Q)\big)$$

with the lexicographic order that makes $\langle i,x\rangle\le_{P+_oQ}\langle j,y\rangle$ iff

$$i<j\lor[i=j=0\land x\le_P y]\lor[i=j=1\land x\le_Qy]\;.$$

(There’s a typo in my copy of the book, which has $U$ and $V$ instead of $P$ and $Q$ for the subscripts on the orders in the displayed line.) Note that although you didn’t give mention it in your question, this definition does mean that every point of $P$ precedes every point of $Q$, so $P+_o[0,1)$ essentially also just adds a new element at the ‘top’ of $P$.

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  • $\begingroup$ That's was my problem. I took by mistake the definition of chapter 2. So silly! Hahahaha $\endgroup$
    – MaríaCC
    Jan 10 '15 at 1:37
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So, I think the exercise can be solve this way:

By definition we have

$$[0,1)=\{x\in \mathbb{N} \, : \, 0 \leq x < 1\}= \{0\}$$

Let $t_P$ be the unique point that satisfies $t_P\in \mathrm{Succ(P)}$ and $t_P\notin P$, and let $\pi:{\mathrm{Succ}(P)}\longrightarrow{P+_o [0,1)}$ defined by $$\pi(p) = \left\{\begin{array}{ll} (1,0) & \text{if } p=t_P \\ (0,p) & \text{otherwise.} \end{array} \right.$$

This map is bijective, because $\pi(\pi^{-1}((a,b)))=(a,b)$ and $\pi^{-1}(\pi(c))=c$ for all $(a,b)\in P+_o [0,1)$, $c\in \mathrm{Succ}(P)$, where $\pi^{-1}$ is:

$$\pi^{-1}((a,b)) = \left\{\begin{array}{ll} b & \text{if } a=0 \\ t_p & \text{if } a=1 \end{array} \right.$$

Let's prove that it is order preserving. If $p_1\leq_{\mathrm{Succ}(P)}p_2$, we have three different cases:

  • If $p_1\leq_P p_2$, then $$\pi(p_1)=(a_1,b_1)=(0, p_1)\leq_{P+_o [0,1)} (0, p_2)=(a_2,b_2)=\pi(p_2),$$ because $a_1=a_2=0$ and $b_1\leq_P b_2$.

  • If $x\in P p_2=t_p$, then $$\pi(p_1)=(a_1,b_1)=(0, p_1)\leq_{P+_o [0,1)} (1,0)=(a_2,b_2)=\pi(p_2),$$ because $a_1<a_2$.

  • If $p_1= p_2=t_p$, then $$\pi(p_1)=(a_1,b_1)=(1,0) \leq_{P+_o [0,1)} (1,0)=(a_2,b_2)=\pi(p_2).$$

So, $\pi$ is a similarity.

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