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Curious Facts about Ordinal Numbers

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