Limit ordinal preserved under arithmetic operations

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I want to check the following statements:

  1. $\alpha+\beta$ is a limit ordinal if and only if $\beta$ is a limit ordinal.
  2. $\alpha\cdot\beta$ is a limit ordinal if and only if either $\alpha$ or $\beta$ is a limit ordinal.

A limit ordinal is an ordinal with no predecessor (consider $0$ a limit).

  1. Not true as $1+0 = 1$ is not a limit?
  2. Multiplying two non-limits we get $\alpha^+\cdot\beta^+=\alpha^+\cdot\beta+\alpha^+$ with predecessor $\alpha^+\cdot\beta+\alpha$. Conversely, I understand the proof that $\alpha\beta$ is a limit if $\beta$ is a limit (they require $\beta$ to be nonzero but zero is not an issue anyway). However, I'm not sure how to prove that $\alpha\beta$ is a limit if $\alpha$ is a limit.
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The first statement is true if you replace ‘if and only if $\beta$ is a limit ordinal’ by ‘if and only if $\beta$ is a non-zero limit ordinal’.

For your other question, suppose that $\alpha$ is a limit ordinal and and $\beta$ is not. Then $\beta=\gamma+1$ for some ordinal $\gamma$, and

$$\alpha\cdot\beta=\alpha\cdot(\gamma+1)=\alpha\cdot\gamma+\alpha\;.$$

If $\alpha\ne 0$ the result follows from (1), and there is clearly no problem if $\alpha=0$.