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

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 \(\omega \notin \omega\).

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:

\(\omega + \omega \neq \omega\)

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.

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*