Cantor normal form multiplication

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SUMMARY

The discussion centers on the multiplication of ordinals in Cantor normal form, specifically addressing the expression \( a \omega = \omega^{\beta_1 + 1} \). The user seeks a proof for this statement, which is derived from the properties of ordinals and their representation in Cantor normal form. The key insight is that multiplying each term \( \omega^{\beta_i} c_i \) by \( \omega \) results in \( \omega^{\beta_{i+1}} \), leading to the conclusion that the largest exponent is \( \beta_1 + 1 \).

PREREQUISITES
  • Understanding of Cantor normal form for ordinals
  • Familiarity with ordinal arithmetic
  • Knowledge of properties of exponents in set theory
  • Basic concepts of ordinal sequences and their ordering
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  • Study the properties of ordinal multiplication in set theory
  • Explore proofs related to Cantor normal form
  • Learn about ordinal exponentiation and its implications
  • Investigate the role of descending sequences in ordinal arithmetic
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Mathematicians, logicians, and students studying set theory or ordinal numbers, particularly those interested in the properties and operations of ordinals in Cantor normal form.

daniel_i_l
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Let's say that a is an ordinal and it's cantor normal form is:
[tex]a = {\omega^{\beta_1}}c_1 + {\omega^{\beta_2}}c_2 + ...[/tex]
I read that
[tex]a \omega = {\omega^{\beta_1+1}}[/tex]
But I couldn't find a proof anywhere.
Can someone give me a source or point me in the right direction so that I can prove it myself?
Thanks.
 
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The [tex]c_{i}'s[/tex] are positive integers, so [tex]c_{i}\omega=\omega[/tex] and:

[tex]\left(\omega^{\beta_{i}}c_{i}\right)\omega=\omega^{\beta_{i}}\omega<br /> <br /> = \omega^{\beta_{i+1}}[/tex]

But the [tex]\beta_{i}[/tex] are in descending order, so their sum is equal to the largest element, which is [tex]\omega^{\beta_{1}+1}[/tex]
 

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