Calculus

Set Theory

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Ordinal Numbers

Solved Problems

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Example 1

Show that the right distributive law \[\left( {\alpha + \beta } \right) \cdot \gamma = a \cdot \gamma + \beta \cdot \gamma \] does not hold for all ordinals.

Example 2

Show that the following rule does not hold for all ordinals: \[{\text{If } \alpha + \gamma = \beta + \gamma,\;}\kern0pt{\text{then } \alpha = \beta .}\]

Example 3

Show that the following rule does not hold for all ordinals: \[{\left( {\alpha \cdot \beta } \right)^2} = {\alpha ^2} \cdot {\beta ^2}.\]

Example 4

Show that the following rule does not hold for all ordinals:

\(\text{If } \gamma \gt 0 \text{ and } \alpha \cdot \gamma = \beta \cdot \gamma,\;\) \(\text{then } \alpha = \beta .\)

Example 5

Simplify the ordinal expression \[\left( {1 + 2\omega} \right)\left( {3 + 4\omega + {\omega^2}} \right).\]

Example 1.

Show that the right distributive law \[\left( {\alpha + \beta } \right) \cdot \gamma = a \cdot \gamma + \beta \cdot \gamma \] does not hold for all ordinals.

Solution.

We know that \(2 \cdot \omega = \omega.\) Then we have

\[2 \cdot \omega = \left( {1 + 1} \right) \cdot \omega = \omega \ne 1 \cdot \omega + 1 \cdot \omega = \omega + \omega.\]

Example 2.

Show that the following rule does not hold for all ordinals: \[{\text{If } \alpha + \gamma = \beta + \gamma,\;}\kern0pt{\text{then } \alpha = \beta .}\]

Solution.

Let \(\alpha = 1,\) \(\beta = 2,\) \(\gamma = \omega.\) We see that

\[\alpha + \gamma = 1 + \omega = \omega,\;\;\beta + \gamma = 2 + \omega = \omega,\]

so \(\alpha + \gamma = \beta + \gamma = \omega,\) though

\[\alpha = 1 \ne \beta = 2.\]

Example 3.

Show that the following rule does not hold for all ordinals: \[{\left( {\alpha \cdot \beta } \right)^2} = {\alpha ^2} \cdot {\beta ^2}.\]

Solution.

Let \(\alpha = \omega\) and \(\beta = 2.\) Check that

\[\left( {\omega \cdot 2} \right)^2 \ne {\omega^2} \cdot {2^2}.\]

Ordinal multiplication is associative. Since \(n \cdot \omega = \omega,\) the left-hand side (\(LHS\)) of this expression is written as

\[{\left( {\omega \cdot 2} \right)^2} = {\left( {\omega \cdot 2} \right)^{1 + 1}} = \left( {\omega \cdot 2} \right) \cdot \omega \cdot 2 = \omega \cdot \left( {2 \cdot \omega} \right) \cdot 2 = \omega \cdot \omega \cdot 2 = {\omega^2} \cdot 2.\]

We see that \(LHS \ne RHS.\)

Example 4.

Show that the following rule does not hold for all ordinals:

\(\text{If } \gamma \gt 0 \text{ and } \alpha \cdot \gamma = \beta \cdot \gamma,\;\) \(\text{then } \alpha = \beta .\)

Solution.

Let \(\alpha = 2,\) \(\beta = 3,\) and \(\gamma = \omega.\) For these ordinal numbers, we have

\[\alpha \cdot \gamma = 2 \cdot \omega = \omega,\;\;\beta \cdot \gamma = 3 \cdot \omega = \omega.\]

So \(\alpha \cdot \gamma = \beta \cdot \gamma = \omega,\) but

\[\alpha = 2 \ne \beta = 3. \]

Example 5.

Simplify the ordinal expression \[\left( {1 + 2\omega} \right)\left( {3 + 4\omega + {\omega^2}} \right).\]

Solution.

We denote this expression by \(\alpha.\) Given that \(n \cdot \omega = \omega,\) we can substitute \(4\omega = \omega\) and \(2\omega = \omega:\)

\[\alpha = \left( {1 + 2\omega} \right)\left( {3 + 4\omega + {\omega^2}} \right) = \left( {1 + \omega} \right)\left( {3 + \omega + {\omega^2}} \right).\]

Now recall that for limit ordinals, \(n + \omega = \omega.\) Therefore,

\[\alpha = \left( {1 + \omega} \right)\left( {3 + \omega + {\omega^2}} \right) = \omega\left( {\omega + {\omega^2}} \right).\]

Using left distributivity and the exponentiation property \({\omega^{\beta + \gamma }} = {\omega^\beta } \cdot {\omega^\gamma },\) we get

\[\alpha = \omega\left( {\omega + {\omega^2}} \right) = \omega \cdot \omega + \omega \cdot {\omega^2} = {\omega^2} + {\omega^3}.\]
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