Ordinal Arithmetic: $n+\alpha=\alpha$

Question

Let $\alpha$ be an ordinal. Show that $n+\alpha=\alpha$.

Proof: Let $W$ be the set with ordinal equal to $\alpha$. Let $A_n=\{a_1,a_2,\cdots, a_n\}$ be the set disjoint from $W$ with its ordinal equal to $n$. We need to show that $A_n+W$ is order-isomorphic to $W$. Let $w_0$ be the minimum element of the set $(W,\le)$. Having defined $w_n$, let $$w_{n+1}=\text{First element of} \ W\setminus\{w_0,w_1,\cdots,w_{n}\}$$

Note that $w_i$'s are all successor elements in $W$. Now define $f:A_n+W\to W$ as follows: $$f(x)=\left\{\begin{array}{ll} w_{k-1} & \mbox{if} \ x=a_k \\ w_{n+k} & \mbox{if} \ x=w_k \\ x & \mbox{otherwise} \end{array} \right. $$

Clearly $f$ is a bijection. Now, suppose $u<v\in A_n+W$. If $u,v\in A_n$, $f(u)<f(v)$. If $u\in A_n, v\in W$ then also $f(u)<f(v)$. Let $u,v\in W$ with $u<v$. If $u,v\notin \{w_k\}$ then $f(u)<f(v)$. If both of them are in $\{w_k\}$ then also the ordering holds. Let $u=w_k$ and $v$ is not in the sequence and $u<v$. Now, $f(u)=w_{k+1}$. Since $v>w_k$ and there is no element between $w_k$ and $w_{k+1}$ and $v$ is outside the sequence, we are forced to conclude that $v>w_{k+1}$. Hence $f(u)=w_{k+1}<v=f(v)$. Thus $A_n+W$ and $W$ are order-isomorphic.

Is this proof okay?

Answer 1

This proof is OK, but it might be helpful to make the things you are assuming (that $n$ is a natural number and $\alpha$ is infinite) more explicit.

Alternatively, you could use the inductive definition of addition instead of the order-theoretic one, the proof might get a little bit shorter:

Let $n < \omega$ be arbitrary. To prove that for every $\alpha \geq \omega$, we have $n + \alpha = \alpha$ by induction, we have to prove the following three things:

• Base case: $n + \omega = \omega$. The inductive definition of addition says that since $\omega$ is a limit ordinal, $n + \omega$ is the limit ordinal of $\{n + \delta : \delta < \omega\} = \{n, n+1, n+2, \dots\}$. Thus we have $n + \omega = \omega$.
• Successor case: For every $\alpha \geq \omega$, if $n + \alpha = \alpha$, then $n + (\alpha + 1) = \alpha + 1$. This follows directly from the inductive definition of addition.
• Limit case: If $\alpha > \omega$ is a limit ordinal and for every $\omega \leq \beta < \alpha$, we have $n + \beta = \beta$, then we must have $n + \alpha = \alpha$. By the definition of addition, $n + \alpha$ is the limit of $$\{n + \beta : \beta < \alpha\} = \{n + x : x < \omega\} \cup \{n + \beta : \omega \leq \beta < \alpha\} \\ = \{n + x : x < \omega\} \cup \{\beta : \beta < \alpha\}$$ The limit of this set is $\alpha$.