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