Inequality involving ordinals

Question

Recently, I came across a difficult question in my set theory module:

Let $\alpha$ be an infinite ordinal. Suppose A ∪ B = $\alpha$, A ∩ B = ∅, and otp(A) = otp(B) = $\beta$. Give an example to show that α may be greater than $\beta + \beta$, where + denotes ordinal addition. In other words, you must give an example of $\alpha$, $\beta$, A, and B satisfying the above conditions such that $\beta + \beta < \alpha$.

With some help, I was able to do this first part of the question. We can let A be $\omega ∪ \{\omega \cdot 2\}$ and B be $[(\omega \cdot 2 + 2)- \omega]- \{\omega \cdot 2\ +1 \}$. Maybe, it is easier for me to say it in words: A will be the set of natural numbers with one additional element with infinite predecessors and this element will be $\omega \cdot 2$. B will be the set of ordinals of the form $\omega + k$, where $k$ is a natural number, and an additional element with infinite predecessors and this element will be $\omega \cdot 2 +1$. So, both sets are of order type $\omega + 1$ because they both are countably infinite and have exactly one element with infinite number of predecessors while the other elements have finitely many predecessors. So, $\beta + \beta = \omega + 1 + \omega + 1 = \omega + \omega + 1 = \omega \cdot 2 + 1 < \alpha = \omega \cdot 2 + 2$.

The above establishes the context for the question I am about to ask. The next part of the question asks us to prove that while it is possible for $\beta + \beta < \alpha$, it is not possible for $\beta + \beta + \beta < \alpha$ for any $\alpha, \beta$. In other words, show that $\beta + \beta + \beta \geq \alpha$ for all $\alpha,\beta$ that satisfy the conditions of the question. I am thinking that we should proceed by ordinal induction on $\beta$ but am kind of stuck. I will appreciate any help on this question. Thanks!

Answer 1

Let us consider the natural sum $\alpha\#\beta$ of two ordinals $\alpha$ and $\beta$ as follows: If $\alpha$ and $\beta$ have the Cantor normal form

• $\alpha = \omega^{\xi_0}\cdot n_0 + \cdots +\omega^{\xi_k}\cdot n_k$,
• $\beta = \omega^{\xi_0}\cdot m_0 + \cdots +\omega^{\xi_k}\cdot m_k $,

for $\xi_0>\cdots \xi_k$ and $n_0,\cdots, n_k,m_0,\cdots m_k<\omega$, then $$ \alpha\#\beta = \omega^{\xi_0}\cdot (n_0+m_0) + \cdots +\omega^{\xi_k}\cdot (n_k+m_k).$$

The following result by de Jongh and Parikh characterize the structural property of the natural sum:

Theorem. (de Jongh and Parick) $\alpha\#\beta$ is the supermum of all possible ordertype of the linear extension of a partial order $\alpha\sqcup \beta$, where $\alpha\sqcup \beta$ is a partial order given by not comparing elements in different sets.

Note that every linear extension of $\alpha\sqcup \beta$ is a well-order: It is clear when both of $\alpha$ and $\beta$ are finite, and if not, fix an infinite decreasing sequence over $L$ and use pigeonhole principle to choose an infinite decreasing sequence over $\alpha$ or $\beta$.

The proof of de Jongh-Parick theorem is in Theorem 3.4 of their paper

D.H.J de Jongh and Rohit Parikh, Well-partial orderings and hierarchies, Indagationes Mathematicae 80 (3), (1977) 195-207.

Going back to your problem, we can see that $\alpha\le \beta\#\beta$. Thus it suffices to prove $\beta\#\beta\le \beta\cdot 3$. Suppose that we have $$\beta = \omega^{\xi_0}\cdot m_0 + \cdots +\omega^{\xi_k}\cdot m_k $$ for $\xi_0>\cdots >\xi_k$ and $m_0,\cdots, m_k<\omega$, $m_0\neq 0$. Then

• $\beta\#\beta = \omega^{\xi_0}\cdot (m_0\cdot 2) + \omega^{\xi_1}\cdot (m_1\cdot 2)+ \cdots +\omega^{\xi_k}\cdot (m_k\cdot 2)$,
• $\beta\cdot 3 = \omega^{\xi_0}\cdot (m_0\cdot 3) + \omega^{\xi_1}\cdot m_1 + \cdots + \omega^{\xi_k}\cdot m_k$.

We can see that when we compare the size two Cantor normal form, it suffices to compare their coefficients in the lexicographical order. This shows $\beta\#\beta<\beta\cdot 3$.