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Let $\eta$ be the order type of $\mathbb{Q}$.

I'm trying to calculate $(1+ \eta) \cdot (\eta + 1)$ and $(\eta + 1) \cdot (1+ \eta)$. So for the first one I'm thinking that you just do this $(1+\eta) \cdot (\eta + 1) = (1+ \eta) \cdot \eta + (\eta + 1) \cdot 1= \eta + \eta + 1= \eta +1$.

Was wondering is this correct. It's hard to think about ordinal arithmetic. I was under the impression you think about it with the washing line method. So left side of the multiplication is the washing line and the right hand side is the clothes.

It's just $(1+\eta) \cdot (1+ \eta)= (1+\eta) \cdot 2$ by similar reasons. However, I this can't be possible. Or maybe I'm doing something wrong.

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Remember that when you multiply order types, you have the left but not the right distributive law. Thus, $(1+\eta)\cdot(\eta+1)=(1+\eta)\cdot\eta+1+\eta$, and $(\eta+1)\cdot(1+\eta)=\eta+1+(\eta+1)\cdot\eta$. For any further simplification, we must see what $(1+\eta)\cdot\eta$ and $(\eta+1)\cdot\eta$ look like.

Suppose that $\langle X,\preceq\rangle$ is a linear order of type $(1+\eta)\cdot\eta$. Then by the definition of $(1+\eta)\cdot\eta$ we can decompose $X$ into pairwise disjoint sets $X_q$ for $q\in\Bbb Q$ in such a way that

  1. if $p,q\in\Bbb Q$ with $p
  2. each $X_q$ is ordered in type $1+\eta$ by $\preceq$ and hence is order-isomorphic to $\Bbb Q\cap[0,1)$ with its usual order.

Note that each $X_q$ is densely ordered by $\preceq$: if $x,y\in X_q$ with $x\prec y$, there is a $z\in X_q$ such that $x\prec z\prec y$.

Clearly $X$ is countably infinite. Now let $x\in X$ be arbitrary; $x\in X_q$ for some $q\in\Bbb Q$. Choose any $y\in X_{q-1}$ and $z\in X_{q+1}$ and note that $y\prec x\prec z$, so that $x$ is neither a first or last element of $\langle X,\preceq\rangle$. Since $x\in X$ was arbitrary, $\langle X,\preceq\rangle$ has no first or last element.

Finally, let $x,y\in X$ with $x\prec y$; there are $p,q\in\Bbb Q$ such that $x\in X_p$ and $y\in X_q$. Clearly $p\le q$. If $p

Thus, $\langle X,\preceq\rangle$ is a dense linear order without endpoints and is therefore order-isomorphic to $\Bbb Q$; that is, it has order type $\eta$.

We can now conclude that $(1+\eta)\cdot(\eta+1)=(1+\eta)\cdot\eta+1+\eta=\eta+1+\eta$. Now the set $\Bbb Q^-$ of negative rationals has order type $\eta$, as does the set $\Bbb Q^+$ of positive rationals, so $\Bbb Q=\Bbb Q^-\cup\{0\}\cup\Bbb Q^+$ has order type $\eta+1+\eta$, and $\eta+1+\eta=\eta$. (Or you could argue as before that $\eta+1+\eta$ is a countable dense order type without endpoints.) The final conclusion is that $$(1+\eta)\cdot(\eta+1)=\eta\;.$$

That was pretty detailed, and the other one is quite similar, so I’ll leave it to you.

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    In the argument that $(1+\eta)\cdot \eta$ is order-isomorphic to $\mathbb Q$ is it fair to say the only threat to it being order-isomorphic to $\mathbb Q$ is the fact that $1+\eta$ has a first element, but given an element of the form $1 \cdot x$ there is a predecessor (for example $1 \cdot (x - \epsilon)$) and then the set is dense below $1 \cdot x$?2012-12-05
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    @Ross: In thinking it out for myself (rather than writing up an explanation) I would indeed focus on the first element of $1+\eta$ as the only possible source of difficulty, but I’d think about the rest a bit differently: that the product has no endpoints is immediate from the fact that the second factor has none, so denseness is the only thing that I really need to worry about. That’s clear within each $1+\eta$ block, and the denseness of the second factor ensures that the $1$ element of such a block has no immediate predecessor.2012-12-05
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    Thanks. I think I'm just confused with the last bit. So for example in $\eta + 1 + \eta$ I can see why this is equal to $\eta$. As you have infinity in either side and an element in the middle. However, I'm confused on say $1+ \eta + \eta$. Should this be $1+ \eta$. As you have a first point followed by two infinities.2012-12-05
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    @simplicity: $\eta+\eta$ is again the type of a countable dense linear order without endpoints, so it’s equal to $\eta$, and $1+\eta+\eta=1+\eta$. That, however, is **not** equal to $\eta$, because it as a left endpoint. It may help to realize that there are exactly four distinct types of countable dense linear orders: $\eta$ (no endpoint), $\eta+1$ (right endpoint), $1+\eta$ (left endpoint), and $1+\eta+1$ (both endpoints).2012-12-05