# For what ordinals does $1+\alpha=\alpha$ hold? For what ordinals does $2\alpha=\alpha$ hold?

For what ordinals does $1+\alpha=\alpha$ hold?

It obviously does not hold for naturals, (though I don't know how to prove it). My proposal is that it holds for any ordinal $\ge\omega$. Assume the contrary: $\exists \gamma \ge \omega$ such that $1+\gamma\ne\gamma$. Let's divide gamma by omega: $\gamma=\omega\tau+\rho$. So, $$1+\omega\tau+\rho\ne\omega\tau+\rho$$

$$\omega\tau+\rho\ne\omega\tau+\rho,$$ which is a contradiction.

For what ordinals does $2\alpha=\alpha$ hold?

My proposal is that it (1) holds for limit ordinals and (2) does not hold for non-limit ordinals. Assume the contrary (1): $\exists$ a limit ordinal $\gamma$ such that $2\gamma\ne\gamma$. Let's divide gamma by omega: $\gamma=\omega\tau+0$. So, $$2(\omega\tau+0)\ne\omega\tau+0$$ $$(2\omega)\tau\ne\omega\tau$$ $$\omega\tau\ne\omega\tau,$$ which is a contradiction.

Assume the contrary (2): $\exists$ a non-limit ordinal $\gamma+k, k>0$ such that $2(\gamma+k)=\gamma+k$. So, $$2\gamma+2k=\gamma+k$$ $$\gamma+2k=\gamma+k$$ $$2k=k,$$ which is a contradiction.

I'd like to know whether my proofs are correct or not, and if they are, whether they're rigourous enough. It would also be great to see alternative ways of proving these.

• what did you do in the second line of the second question? Commented May 30, 2018 at 18:36
• @GuillermoMosse I am sorry, could you please specify the line? I can't notice anything that may need clarification. (Sorry!) Commented May 31, 2018 at 20:10
• You have an error when you go from $2\gamma + 2k = \gamma + k$ to $\gamma + 2k = \gamma + k$ Commented Jun 1, 2018 at 18:00
• @GuillermoMosse Why? (Implied that $\gamma$ is a limit ordinal and $k>0$ is a natural.) Commented Jun 2, 2018 at 3:51
• oh, then it's ok! sorry Commented Jun 4, 2018 at 17:49

It seems that those questions (124, and 125) are from A. Shen's "Basic Set Theory", which I've encountered recently.

You proofs help me about clarity in my approach, but I think the proof should be used the materials which are provided ealier in the book. It seems that you are using the result of Problem 126 (limit ordinals can be represented as $$\omega\cdot\alpha$$), it's not wrong but if (like myself) studying from Shen's book, the problem 126 must be proved first.

So I try to modified your proof using "known" material provide in the book, which does NOT mean my proof is correct, but nevertheless I post as an answer instead of new question:

• Problem 1:

If $$\alpha<\omega$$, then the sets of order type $$\alpha$$ are finite sets, and have finite cardinality. We have $$\left|A\right|<1+\left|A\right|$$, implies there is no one-to-one correspondence between the sets of order type $$\alpha$$ and $$1+\alpha$$.

Consider an ordinal $$\alpha$$ that $$\geq\omega$$.

We have $$1+\omega=\omega$$.

Suppose that $$\forall\beta\left(\omega\leq\beta<\alpha\right)\rightarrow\left(1+\beta=\beta\right)$$.

If $$\alpha$$ is a nonlimit ordinal, we have $$\alpha=\beta+1$$. Let $$A,A^{\prime}$$ are sets have order types $$\alpha$$ and $$1+\alpha$$, and let $$a$$ is the largest element of A (and also, of $$A^{\prime}$$). We have $$A\setminus\left\{ a\right\} ,A^{\prime}\setminus\left\{ a\right\}$$ have order type $$\beta$$ and $$1+\beta$$. By induction hypothesis, $$A\setminus\left\{ a\right\}$$ is isomorphic to $$A^{\prime}\setminus\left\{ a\right\}$$ by a map $$f$$. Let $$f\left(a\right)=a$$, we have isomorphism between $$A$$ and $$A^{\prime}$$.

Consider $$A$$ has order type $$\alpha$$, we have $$\left\{ b\right\} \cup A$$ has order type $$1+\alpha$$.

Consider an initial segment $$I$$ of $$\left\{ b\right\} \cup A$$, that $$I\neq\left\{ b\right\}$$ , let $$a$$ (as $$I\neq\left\{ b\right\}$$ , implies $$a\neq b$$ (or $$a\in A$$)) is the least element of $$\left(\left\{ b\right\} \cup A\right)\setminus I$$, we have $$\forall x\in I, x. While $$b$$ is the least element of $$\left\{ b\right\} \cup A$$, we have $$\forall x\in I\setminus\left\{ b\right\} ,x, implies that $$I\setminus\left\{ b\right\}$$ is an initial segment of $$A$$, and $$I\setminus\left\{ b\right\} =\left[0,a\right)$$. (that part, I feel hard to explain clearly).

That is if an initial segment of $$1+\alpha$$ has the form $$1+\beta$$, where $$\beta$$ is an initial segment of $$\alpha$$.

If $$\alpha$$ is a limit ordinal. Suppose that $$1+\alpha\neq\alpha\rightarrow1+\alpha>\alpha$$, implies that $$\alpha$$ is isomorphic to an initial segment $$1+\beta$$ of $$1+\alpha$$, where $$\beta$$ is an initial segment of $$\alpha$$. By induction hypothesis, $$\beta=1+\beta$$, which means $$\alpha$$ is isomorphic to its initial segment $$\beta$$. By Theorem 21 (of the book), it's either impossible or $$\beta=\alpha$$, but then contradicts to our hypothesis that $$1+\alpha\neq\alpha$$.

So, we have for any $$\alpha\geq\omega, 1+\alpha=\alpha$$.

• Problem 2:

Suppose that $$\alpha$$ is a nonlimit ordinal, then there's an ordinal $$\beta$$, $$\beta+1=\alpha$$,

$$2\cdot\left(\beta+1\right) =\beta+1$$ $$2\cdot\beta+2 =\beta+1$$

Consider $$A$$ and $$B$$ have order types $$\alpha$$ and $$2\cdot\alpha$$, so $$A$$ is (ordered) isomorphic to $$B$$. This means the greatest element of $$A$$ corresponds to the greatest element of $$B$$, implies that $$2\cdot\beta+1=\beta$$. But we have $$\beta\leq2\beta<2\beta+1$$, so there doesn't exist nonlimit ordinal $$\alpha$$ that $$2\alpha=\alpha$$.

Suppose that $$\alpha$$ is a limit ordinal, and $$2\alpha\neq\alpha\rightarrow2\alpha>\alpha$$. As we know (p.89) that if $$\alpha<2\alpha$$, then there's unique representation of $$\alpha$$, $$\alpha=2\alpha_{1}+\beta$$, where $$\alpha_{1}<\alpha, \beta<2$$.

If $$\beta=1$$, it contradicts to hypothesis that $$\alpha$$ is a limit ordinal.

If $$\beta=0, \alpha=2\alpha_{1}$$, where $$\alpha_{1}$$ is also limit ordinal. If $$2\alpha_{1}=\alpha_{1}\rightarrow\alpha=\alpha_{1}=2\alpha_{1}=2\alpha$$.

If $$2\alpha_{1}\neq\alpha_{1}$$, and $$\alpha_{1}=2\alpha_{2}$$..., consider the set of all $$\alpha_{i}$$, that $$2\alpha_{i}\neq\alpha_{i},\alpha_{i}=2\alpha_{i+1}$$. The set must contain the least element $$\alpha_{j}$$, that is $$2\alpha_{j}\neq\alpha_{j}$$, apply the same argument above, it must $$\alpha_{j}=2\alpha^{\prime}\rightarrow2\alpha_{j}>2\alpha^{\prime}\rightarrow\alpha_{j}>\alpha^{\prime}$$. Because $$\alpha_{j}$$ is the least element of the set, we have $$\alpha^{\prime}$$ doesn't belong to the set, or $$2\alpha^{\prime}=\alpha^{\prime}\rightarrow\alpha^{\prime}=\alpha_{j}$$, which contradicts.

(I feel this part is over-complicated).

• Busy with college and having no spare time to read and understand your answer, I still have to express my gratitude for praising my question with attention in a form of such a long post. Yes, that book was The One. (Да, у меня тоже книга Шеня). Commented Nov 19, 2018 at 13:23
• @fragileradius: Note that the theorem and the page's number I mentioned here are from English version of the book.
– duqu
Commented Nov 20, 2018 at 7:42