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suppose A is a linear order set with a copy of rationals in it that is $A=B\cup\{\bar{r}:r\in\mathbb{Q}\cap[0,1]\}$. is there an orde preserving map that preservs sup and inf betwwen A and $[0,1]$ in witch the image of $\bar{r}$ is $r$. for $[0,1]^2$ in witch the image of $\bar{r}$ is $(r,r)$ or $(0,r)$ who?

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closed as off topic by Goldstern, Ricky Demer, Pietro Majer, Asaf Karagila, Andy Putman Dec 2 '12 at 19:44

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Do you mean "preserving $<$" (in this case the answer is obviously no) or "preserving $\le$" (in this case the answer is yes). –  Goldstern Dec 2 '12 at 9:17
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Consider the rational order with two $\frac12$s, an upper one and a lower one. This order contains a copy of $\mathbb{Q}$, by taking only the lower one, but if you map back into $\mathbb{Q}$, the upper one will have to skip over some rationals. Similar examples arise whenever you replace each rational by any nonempty linear order, at least one of which has more than one point, e.g. $\mathbb{Q}$ copies of $\mathbb{Z}$. –  Joel David Hamkins Dec 2 '12 at 14:03
    
Yes, my previous comment was partially incorrect. –  Goldstern Dec 2 '12 at 16:49
    
Goldstern, I was responding to the OP; what was wrong with your comment? My example and many others show that $\lt$-preservation can be impossible, but by mapping every point $x$ to the sup of the $r$ for which $\bar r\leq x$ we get $x\leq y\implies f(x)\leq f(y)$, preserving $\leq$, infs and sups, and mapping $\bar r\mapsto r$. Right? –  Joel David Hamkins Dec 2 '12 at 16:58
    
@Joel: You are right. @Amin: Your question will be closed soon, mainly because it does not look like a research level problem. Also, you have not given any background or motivation, which makes the question look like homework. –  Goldstern Dec 2 '12 at 17:24
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