Ordinal Addition is Associative

Theorem
Ordinal addition is associative, i.e.:


 * $\paren {x + y} + z = x + \paren {y + z}$

holds for all ordinals $x$, $y$ and $z$.

Proof
By Transfinite Induction on $z$.

Basis for the Induction
This proves the basis for the induction.

Induction Step
Suppose that:


 * $x + \paren {y + z} = \paren {x + y} + z$

Then:

This proves the induction step.

Limit Case
Let $z \in K_{II}$ be a limit ordinal, and suppose that:


 * $\forall w \in z: x + \paren {y + w} = \paren {x + y} + w$

Then it follows that:

By the hypothesis, we have that $z \in K_{II}$.

From Limit Ordinals Preserved Under Ordinal Addition, also $y + z \in K_{II}$.

Therefore, by definition of ordinal addition:


 * $x + \paren {y + z} = \ds \bigcup_{n \mathop \in y + z} \paren {x + n}$

So it is sufficient to prove that:


 * $\ds \bigcup_{w \mathop \in z} \paren {x + \paren {y + w} } = \bigcup_{n \mathop \in y + z} \paren {x + n}$

Take any $w \in z$.

Then by Membership is Left Compatible with Ordinal Addition, $y + w < y + z$.

Setting $n = y + w$, we now have that:


 * $x + \paren {y + w} \le x + n$

By Supremum Inequality for Ordinals, it now follows that:


 * $\ds \bigcup_{w \mathop \in z} \paren {x + \paren {y + w} } \le \bigcup_{n \mathop \in y + z} \paren {x + n}$

To prove the converse inequality, take any $n \in y + z$.

By definition of ordinal addition, this means:


 * $n \in \ds \bigcup_{w \mathop \in z} \paren {y + w}$

and so, for some $w \in z$, we have $n \in y + w$.

By Membership is Left Compatible with Ordinal Addition, this yields:


 * $x + n \in x + \paren {y + w}$

and whence by Supremum Inequality for Ordinals:


 * $\ds \bigcup_{n \mathop \in y + z} \paren {x + n} \le \bigcup_{w \mathop \in z} \paren {x + \paren {y + w} }$

By definition of set equality:


 * $\ds \bigcup_{n \mathop \in y + z} \paren {x + n} = \bigcup_{w \mathop \in z} \paren {x + \paren {y + w} }$

This proves the limit case.

Hence the result, by Transfinite Induction.

Also see

 * Natural Number Addition is Associative/Proof 2