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This question already has an answer here:

Show that $(\omega +3)\cdot\omega=\omega\cdot\omega$.

Is this just $(\omega +3)\cdot\omega=(\omega +\omega)\cdot\omega=\omega\cdot\omega$?

Also, could someone suggest a good book for set theory?

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marked as duplicate by Noah Schweber, gt6989b, Shailesh, Daniel W. Farlow, choco_addicted Jun 9 '16 at 3:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What is $\omega$ here? $\endgroup$ – gt6989b Jun 8 '16 at 19:48
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    $\begingroup$ Book recommendations: math.stackexchange.com/q/1491464/212120 $\endgroup$ – Pedro Sánchez Terraf Jun 8 '16 at 19:54
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    $\begingroup$ @Noah: I take a narrow view of what constitutes a duplicate: I’m not willing to call that one a duplicate unless the OP agrees that it answers the question. $\endgroup$ – Brian M. Scott Jun 8 '16 at 20:34
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    $\begingroup$ One of the best serious set theory texts at the advanced undergraduate/beginning graduate level is Hrbacek & Jech, Introduction to Set Theory, Third Edition, Revised and Expanded. $\endgroup$ – Brian M. Scott Jun 8 '16 at 20:38
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    $\begingroup$ @BrianM.Scott: Generally, I agree with that policy, but it sure seems in this instance that $3$ can very easily be viewed as a special case of $2$. $\endgroup$ – Brian Tung Jun 8 '16 at 22:12
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Note that the following lines follow immediately from the definition of ordinal multiplication $\alpha \cdot \lambda$ for limit ordinals $\lambda$. On the one hand, we have

$ \begin{align*} (\omega + 3) \cdot \omega =& \sup \{ (\omega + 3) \cdot n \mid n < \omega \} \\ \ge& \sup \{ \omega \cdot n \mid n < \omega \} \\ =& \omega \cdot \omega. \end{align*} $

On the other hand

$ \begin{align*} (\omega + 3) \cdot \omega =& \sup \{ (\omega + 3) \cdot n \mid n < \omega \} \\ \le& \sup \{ (\omega + \omega) \cdot n \mid n < \omega \} \\ =& \sup \{ \omega \cdot (n+1) \mid n < \omega \} \\ =& \omega \cdot \omega. \end{align*} $

Hence $(\omega + 3) \cdot \omega = \omega \cdot \omega$.

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  • $\begingroup$ I think you have to use normality (or continuity) of the ordinal product to justify last displayed step (i.e., $\sup \{ \omega \cdot (n+1) \mid n < \omega \} = \omega \cdot \omega$.) $\endgroup$ – Pedro Sánchez Terraf Jun 9 '16 at 0:23
  • $\begingroup$ @PedroSánchezTerraf $\{\omega \cdot (n+1) \mid n < \omega \} = \{\omega \cdot n \mid 0 < n < \omega \} \subseteq \{ \omega \cdot n \mid n < \omega \}$. Hence $\sup \{ \omega \cdot (n+1) \mid n < \omega \} \le \omega \cdot \omega$ and monotonicity actually yields equality. $\endgroup$ – Stefan Mesken Jun 9 '16 at 7:10
  • $\begingroup$ That's perfect; the point I wanted to make is that you need to use a bit more than the bare definition of product. $\endgroup$ – Pedro Sánchez Terraf Jun 9 '16 at 11:07
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    $\begingroup$ @LeAnhDung Yep, that's it. $\endgroup$ – Stefan Mesken Dec 1 '18 at 11:37
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    $\begingroup$ @LeAnhDung You're most welcome! $\endgroup$ – Stefan Mesken Dec 1 '18 at 11:38
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By using only that the ordinal product is associative, non-decreasing in each variable, and the fact that $2\cdot\omega=\omega$, you can make your argument formal.

Since $\cdot$ is monotonic, $$ (\omega +3)\cdot\omega\geq\omega\cdot\omega. $$ By the same reason, $$ (\omega +3)\cdot\omega\leq(\omega +\omega)\cdot\omega=(\omega\cdot 2)\cdot\omega= \omega\cdot(2\cdot\omega), $$ where the last equality follows by associativity. Finally, since $2\cdot\omega=\omega$, we obtain $$ (\omega +3)\cdot\omega\leq\omega\cdot(2\cdot\omega) = \omega\cdot\omega. $$ Hence we have both inequalities.

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\begin{align*}(\omega+3)+(\omega+3)+(\omega+3)\dots&=\omega+(3+\omega)+(3+\omega)+(3+\omega)+(3+\omega)\dots \\ &=\omega+\omega+\omega+\omega+\dots \\ &=\omega\cdot\omega \end{align*}

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  • $\begingroup$ I only used associativity. $\endgroup$ – Jacob Wakem Jun 9 '16 at 0:09
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    $\begingroup$ And infinite sums... $\endgroup$ – Pedro Sánchez Terraf Jun 9 '16 at 11:08

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