Operations on ordinal numbers

Question

1. Let $w$ be an ordinal for a denumerable set. Prove that$(w+w)w=ww$

2. Let $A$ and $B$ be sets. Let $A$ be ordered by $G$ and $B$ by $H$. Let $f$ be an isomorphism such that $x≦y$ in $G$ implies $f(x)≦f(y)$ in $H$. Now, reorder $A$ by an order relation $G'$. Then does there exist an isomorphism $f'$ such that $x≦y$ in $G'$ implies $f'(x)≦f'(y)$ in $H$?

Dear Asaf

I think the book I'm studying is not really a good one. I'm really sorry that everytime i come up with easy questions, seems like I'm using this website to just do my homework quickly, but i don't.. I want you to know that I'm studying set theory by myself and this book gives only some definitions and leaves important theorems in exercises. I swear that i post questions I've tried to solve at least for an hour or a day.. Plus, even though i have solved problems, for some problems, I didn't like the way I solved because it's kinda messy so I wanted to know how to solve the problems easily

Answer 1

The result is false. A counterexample is obtained by taking $w=\omega^2+\omega+1$.

Note $w+w=\omega^2\cdot2+\omega+1$. Now, adding $z=\omega^2\cdot2+\omega+1$ to itself $\omega$ times is $\omega^3$, and adding $z$ to itself $\omega^2$ times is therefore $\omega^4$.

Then $(w+w)w=z(\omega^2+\omega+1)=\omega^4+\omega^3+\omega^2\cdot2+\omega+1$.

On the other hand, $ww=\omega^4+\omega^3+\omega^2+\omega+1<(w+w)w$.

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In general, one can obtain many counterexamples by considering $w$ whose Cantor normal form includes indecomposables of several kinds, not all of them limit ordinals.

Answer 2

I assume that by $w$ you mean $\omega$.

Then showing $\omega\omega=(\omega+\omega)\omega$ is relatively easy.

The first one is the set $\omega\times\omega$, the second one is $(\omega\times\{0\}\cup(\omega\times\{1\}))\times\omega$, both of them with antilexicographic order, i.e. the second coordinate is more important.

The isomorphism between the two sets is $(a,2b)\mapsto((a,0),b)$ and $(a,2b+1)\mapsto((a,1),b)$.

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In fact, this can be visualized quite easily $\alpha\times\beta$ means that you take ordinal $\beta$ and replace each element by a copy of $\alpha$. I.e. $\alpha\beta$ consists of "$\beta$-many" copies of $\alpha$ which you put one after each another, and they are ordered according to $\beta$.

In this sense $\omega\omega$ consists of $\omega$ copies of $\omega$.

What is $(\omega+\omega)\omega$? I've put several pairs of copies of $\omega$ after each other. This is the same as $2\omega$ copies of $\omega$, but $2\omega=\omega$. (If I replace each element in $\omega$ by a pair of elements, then I do not change the order type.)

At the moment I do not have time to draw a nice picture illustrating this but this might help too:

• $\omega$ looks like $\ast \ast \ast \cdots$
• $2$ is simply the two points $(\circ \circ)$
• To get $2\omega$ we simply replace each $\ast$ by $(\circ \circ)$ and we get $(\circ \circ) (\circ \circ) (\circ \circ) \cdots$
• Of course we can take parentheses away (they are just an auxiliary thing to show where the copies of $2$ were added), so this is simply $\circ \circ \circ \circ \circ \circ \cdots$
• The two things look the same, only we have used $\ast$ to denote the elements in one of them and $\circ$ in another one.

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You can prove the same thing using some rules for cardinal arithmetic, namely you have $\omega+\omega=\omega\cdot 2$ so it suffices to show that $2\omega=\omega$ (which might be easier for you). If you know both these things you get $$(\omega+\omega)\omega=(\omega2)\omega=\omega(2\omega)=\omega\omega.$$

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More importantly I would like to ask you to clarify the second part (perhaps ti would be even better to post it as a separate question).

You have something like this:

Let $A$ and $B$ be sets. Let $A$ be ordered by $\le_G$ and $B$ by $\le_H$. Let $f$ be an isomorphism such that $x\le_G y$ implies $f(x)\le_H f(y)$. Now, reorder $A$ by an order relation $\le_{G'}$. Then does there exist an isomorphism $f'$ such that $x\le_{G'}y$ implies $f'(x)\le_H f'(y)$?

What precisely do you mean under isomorphism? If you mean isomorphism of ordered sets $(A,\le_G)$ and $(B,\le_H)$, then the definition of isomorphism includes that $x\le_G y$ $\Leftrightarrow$ $f(x)\le_H f(y)$. So your sentence such that $x\le_G y$ implies $f(x)\le_H f(y)$ seems to be redundant. Now if there are no conditions on $\le_{G'}$, you are basically asking whether two orders of the same set must be isomorphic. This is certainly not true. (But maybe I've misunderstood the question.)