Posted in: math , philosophy

This builds on previous notes. Some cool facts about ordinals:

The addition of ordinal numbers is not commutative.

We have the following definition for addition on ordinal numbers:

(i) $$\alpha + 0 = \alpha$$

(ii) $$\alpha + \beta’ = (\alpha + \beta)’$$

(iii) if $$\beta$$ is a limit ordinal then $$\alpha + \beta = \bigcup_{\gamma \in \beta}(\alpha + \gamma)$$

It’s easy to see that addition is not commutative with this example:

$$1 + \omega \neq \omega + 1$$

Starting from the left side we have that:

$$1 + \omega = \bigcup_{\gamma \in \omega}(1 + \gamma)$$ which expands to:

$$1 \cup 2 \cup 3 \cup … = \{0\} \cup \{0, 1\} \cup \{0, 1, 2\} \cup … = \omega$$

On the right side we have:

$$\omega + 1 = (\omega + 0)’ = \omega’ = \omega \cup \{\omega\}$$

and $$\omega \neq \omega’$$ simply because $$\omega \in \omega’$$ and following the ordinal definition $$\omega \notin \omega$$.

Ordinal multiplication is not distributive from the right

We have the following definition for multiplication on ordinal numbers:

(i) $$\alpha . 0 = 0$$

(ii) $$\alpha . \beta’ = \alpha . \beta + \alpha$$

(iii) if $$\beta$$ is a limit ordinal then $$\alpha . \beta = \bigcup_{\gamma \in \beta}(\alpha . \gamma)$$

It’s easy to see that multiplication is not distributive from the right with this example:

$$(1 + 1)\omega \neq 1 . \omega + 1 . \omega$$

On the left side of the equation we have:

$$(1 + 1) . \omega = 2 . \omega$$ and we can use (iii) to get

$$2 . \omega = \bigcup_{\gamma \in \omega}(2 . \gamma)$$ which expanded looks like:

$$(2 . 1) \cup (2 . 2) \cup (2 . 3) \cup …$$ which is:

$$\{0, 1\} \cup \{0, 1, 2, 3\} \cup \{0, 1, 2, 3, 4, 5\} \cup … = \omega$$

On the right hand side we have:

$$1 . \omega + 1 . \omega = \omega + \omega > \omega$$ which also means that:

$$\omega + \omega \neq \omega$$

The exponent rule is not always true

We have the following definition for exponentiation on ordinal numbers:

(i) $$\alpha^0 = 1$$

(ii) $$\alpha^{\beta’} = \alpha^\beta . \alpha$$

(iii) if $$\beta$$ is a limit ordinal and $$\alpha > 0$$ then $$\alpha^\beta = \bigcup_{\gamma \in \beta}(\alpha^\gamma)$$

(iv) if $$\beta$$ is a limit ordinal and $$\alpha = 0$$ then $$\alpha^\beta = 0$$

Let’s try to see the counterexample:

$$(\omega . 2)^2 \neq \omega^2 . 2^2$$

If we start from the right side we have:

$$\omega^2 . 2^2 = \omega^2 . 4 = \omega^2 . 3 + \omega^2 = \omega^2 . 2 + \omega^2 + \omega^2$$ and if we continue the expansion we have:

$$\omega^2 + \omega^2 + \omega^2 + \omega^2 > \omega^2 + \omega^2$$ so we know that:

$$\omega^2 + \omega^2 + \omega^2 + \omega^2 \neq \omega^2 + \omega^2$$ let’s call this (1).

If we expand the left side we get:

$$(\omega . 2)^2 = (\omega . 2) . (\omega . 2)$$ since the product of ordinals is associative we can do:

$$\omega . (2 . \omega) . 2$$ let’s call this equation (2).

If we expand the product $$2 . \omega$$ we get:

$$2 . \omega = \bigcup_{\gamma \in \omega}(2 . \gamma)$$ which expands to:

$$(2 . 1) \cup (2 . 2) \cup (2 . 3) \cup … = \omega$$. Going back to (2) we now have:

$$\omega . (2 . \omega) . 2 = \omega . \omega . 2 = \omega^2 . 2 = \omega^2 + \omega^2$$ and by (1) we know this is different to the right hand side.

Things that are undefined in analysis have a clear meaning for ordinals

This might be arguably the coolest thing. So for example $$0^0, \infty^0, 1^\infty$$ have clear meanings:

$$0^0 = \omega^0 = \alpha^0 = 1$$

and

$$1^\omega = 1^\beta = 1$$

Content taken from Axiomatic Set Theory by Patrick Suppes