# dense linear orders DLO

I am asked to prove that if I have two models of dense linear orders DLOs WITH the minimum and maximum. must be izomorpic to each other by fining direct izomorphy.

I seem to always get stuck because the desity i just cannot understand how to count them

im trying something like this: assuming $$M_1=\{a_1 ... a_2 ...\}$$ assuming $$M_1=\{b_1 ... b_2 ...\}$$

I choose $$a_i$$ but if I choose $$H(a_1)=b_i$$

how can I know its in the same order?

sorry for my bad english - translating from hebrew

You are referring to a standard back and forth argument that goes back to Cantor.

The result is that if $A$ and $B$ are densely ordered countable sets with first and last element, then $A$ and $B$ are order-isomorphic.

If $A$ and $B$ are densely ordered with first and last element, and $A$ and $B$ have the same uncountable cardinality, then $A$ and $B$ need not be order isomorphic.

For the proof in the countable case, let $a_s$ be the smallest element of $A$, and let $a_m$ be the biggest. Define $b_s$ and $b_m$ analogously. We will want our isomorphism to take $a_s$ to $b_s$, and $a_m$ to $b_m$. The rest of $A$ (and $B$) is a countable densely ordered set with no first or last element. I am making these obvious comments because you may have been given an isomorphism proof in that case.

Roughly speaking, we build the rest of the isomorphism gradually. List the non-endpoints of $A$ and $B$ as $a_1,a_2,\dots$ and $b_1, b_2,\dots$. Build the isomorphism $\phi$ by first letting $a'_1=a_1$ and $b'_1=b_1$, and putting $\phi(a_1')=b_1'$.

Suppose we have found objects $a'_1,\cdots,a'_n$ and $b'_1,\dots,b'_n$ for which the mapping that takes $a'_i$ to $b'_i$ ($i=1$ to $n$) respects order. If $n$ is odd, look at the smallest $w$ such that $b_w$ is not (yet) one of the $b'_i$. (Note, it is $b_m$, we are working with the arbitrary listing of the elements of $B$, not with the "real" ordering of $B$.)

Let $b'_{n+1}=b_w$. Find any $a\in A$ such that $a$ has the same order relationship to the $a'_i$ as $b_w$ has to the $b'_i$. (There is such an $a$, because our order is dense.) Let $a'_{n+1}=a$.

For $n$ even, do the same sort of thing, except that we use the smallest $w$ such that $a_w$ is not one of the $a'_i$, and let $a'_{n+1}=a_w$. Then we find a suitable $b$ to map $a'_{n+1}$ to.

Very sketchy! The Wikipedia article has more detail.

Hint: start by defining where the minima and maxima must go and then go "back and forth". I'll elaborate if you'd like.

• so can I say that $$H(a_min) =b_min$$ and $$H(a_max) =b_max$$ and than use back and forth like @Nicolas showed? – Nahum Jun 9 '12 at 17:22
• Yes, that's what you need to do. – Quinn Culver Jun 9 '12 at 17:24
• thats horrible! would copy pasting the proof from en.wikipedia.org/wiki/Back-and-forth_method and adding the lines in my last comment will be correct answer? OK thats wrong, they count on the facttheres no maximum – Nahum Jun 9 '12 at 17:25
• Yes, it would be correct, but you should try writing down the proof in your own words! – Alex Kruckman Jun 9 '12 at 17:29
• they tought us nothing alike! at all. not even close :( – Nahum Jun 9 '12 at 17:30