Examples of Ordinal Arithmetic
We’ll use frequently the exercise that ω α + ω β = ω β if α < β.
Lemma 0.1. (ω α0 · k0 + ω α1 · k1 + · · · ω αn · kn ) · β = (ω α0 · k0 · β), if β is a limit
and = (ω α0 · k0 · β) + (ω α1 · k1 + · · · + ω αn · kn ) if β is a successor (T stands for
“tail.”)
Proof. By induction on β. For β = 0 it is trivial (provided we consider 0 a limit
ordinal!; it is also trivial for β = 1 if you prefer to start there). Let α denote
ω α0 · k0 + ω α1 · k1 + · · · ω αn · kn . Let T (for “tail”) denote ω α1 · k1 + · · · + ω αn · kn .
Case I.) β is a successor, say β = γ + 1. Then α · β = α · (γ + 1) = α · γ + α =
(ω α0 ·k0 ·γ)+?T +(ω α0 ·k0 +ω α1 ·k1 +· · · ω αn ·kn ) where ?T denotes T if γ is a successor
and denotes 0 if γ is a limit. From the exercise T + ω α0 · k0 = ω α0 · k0 , so in either
case this becomes = ω α0 ·k0 ·γ +ω α0 ·k0 +T = ω α0 ·k0 ·(γ +1)+T = ω α0 +1 ·k0 ·β +T
which is what is claimed.
Case II.) β is a limit. Then
α · β = α · ( sup β) = sup α · β 0 = sup (ω α0 · k0 · β 0 + T ),
β 0 <β
β 0 <β
β 0 <β
where we have assumed without loss of generality that β 0 is a successor ordinal
(every limit is a limit of successors). Now ω α0 ·k0 ·β 0 +T < ω α0 ·k0 ·β 0 +T +ω α0 ·k0 =
ω α0 · k0 · β 0 + ω α0 · k0 = ω α0 · k0 · (β 0 + 1). Since β 0 + 1 < β whenever β 0 < β (as β
is a limit) the supremum is bounded by supβ 0 <β (ω α0 · k0 · (β 0 + 1)) = ω α0 · k0 · β.
Clearly this is also a lower bound for the supremum, and we are done.
Using this lemma, ordinal multiplication is now straightforward. For example,
here is the example we did in class:
(ω ω
2
+ω
2
· 3 + ω 2 · 3 + 11) · (ω ω · 2 + ω ω · 2 + 17)
= ωω
2
+ω
· 3 · (ω ω · 2 + ω ω · 2 + 17) + ω 2 · 3 + 11
2
= ωω
2
+ω
· (ω ω · 2 + ω ω · 2 + 51) + ω 2 · 3 + 11
= ωω
2
·2
2
· 2 + ωω
2
+ω·2
· 2 + ωω
2
+ω
· 51 + ω 2 · 3 + 11
which is now in Cantor normal form.
Now let’s turn to exponentiation. To warm up, let’s first compute αm where
m ∈ ω.
Lemma 0.2. (ω α0 · k0 + ω α1 · (
k1 + · · · ω αn · kn )m = ω α0 ·m · k0 + ω α0 ·(m−1)+α1 · k1 +
ω α0 ·(m−1)+αn · kn if αn > 0
· · · + ω α0 ·(m−1)+αn−1 · kn−1 +
αm−1 · kn
if αn = 0
Note that if αn = 0 (i.e., α is a successor ordinal), then the last term requires
us to recursively compute αm−1 .
Proof. First assume that αn > 0, so α is a limit. Since αm = αm−1 · α, and α is a
limit ordinal, applying the multiplication rule m − 1 times gives
αm =αm−1 · α = (ω α0 · k0 )m−1 · α = (ω α0 · k0 )m−1 · (ω α0 · k0 + · · · + ω αn · kn ) =
ω α0 ·m · k0 + ω α0 ·(m−1)+α1 · k1 + · · · + ω α0 ·(m−1)+αn · kn
1
2
Next assume that α0 = 0, so that last term is kn and hence α is a successor
ordinal. Let H = ω α0 · k0 + · + ω αn−1 · kn−1 (H stands for “head”). Then αm =
αm−1 · (H + kn ) = αm−1 · H + αm−1 · kn . Since H is a limit ordinal, the first
term, using the multiplication lemma, becomes (ω α0 · k0 ))m−1 · H = ω α0 ·m · k0 +
ω α0 ·(m−1)+α1 · k1 + · · · + ω α0 ·(m−1)+αn−1 · kn−1 and the result follows.
2
Example. Let’s compute (ω ω +ω ·3+ω ω ·7+2)3 . Let α denote ω ω
Using the previous lemma we have:
(ω ω
2
+ω
· 3 + ω ω · 7 + 2)3 = ω (ω
2
+ω)·3
· 3 + ω (ω
= ωω
2
·3+ω
· 3 + ωω
2
·2+ω·2
· 7 + α2 · 2
= ωω
2
·3+ω
· 3 + ωω
2
·2+ω·2
· 7 + ω (ω
=ω
ω 2 ·3+ω
= ωω
2
·3+ω
·3+ω
ω 2 ·2+ω·2
· 3 + ωω
2
·7+ω
·2+ω·2
2
+ω)·2
ω 2 ·2+ω
· 7 + ωω
2
·2+ω
2
+ω)·2+ω
· 3 · 2 + ωω
·6+ω
· 6 + ωω
+ω
·3+ω ω ·7+2.
· 7 + α2 · 2
2
ω 2 +ω·2
2
2
+ω·2
+ω+ω
·7·2+α·4
· 14 + α · 4
· 14 + ω ω
2
+ω
· 12 + ω ω · 28 + 8
Computing αβ when β is a limit ordinal is easier.
Lemma 0.3. If β is a limit, then (ω α0 · k0 + · · · + ω αn · kn )β = ω α0 ·β .
Proof. We prove this by induction on β. For β = ω, note that (ω α0 · k0 + · · · + ω αn ·
kn )m lies between ω α0 ·m and (ω α0 · (k0 + 1))m = ω α0 ·m · (k0 + 1) < ω α0 ·m+1 , and
both of these sup up to ω α0 ·ω .
If β is a limit of limit ordinals, then using induction we have
0
0
(α)β = sup ω β = sup ω α0 ·β = ω α0 ·β
β 0 <β
β 0 <β
β 0 limit
β 0 limit
If β is not a limit of limit ordinals, then β = β 0 + ω for some limit ordinal β 0 .
Then
αβ = αβ
0
+ω
0
0
= αβ · αω = ω α0 ·β · ω α0 ·ω = ω α0 ·β
0
+α) ·ω
= ω α0 ·(β
0
+ω)
= ω α0 ·β .
Since every ordinal β is of the form λ + m for some limit ordinal λ, we now have
complete rules for ordinal exponentiation.
2
Example. We compute αβ where α = ω ω +ω · 3 + ω ω · 7 + 2 (as in the above
example) and β = ω 3 + 3.
3
αβ = αω · α3 = ω (ω
ωω
4
+ω 2 ·3+ω
+ ωω
4
· 3 + ωω
2
+ω +ω
2
4
+ω)·ω 2
4
· α3 = ω ω · α3 =
+ω 2 ·2+ω·2
· 12 + ω ω
4
+ω
· 7 + ωω
4
+ω 2 ·2+ω
· 28 + ω 4 · 8
· 6 + ωω
4
+ω 2 +ω·2
· 14