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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What is $\omega$ here? – gt6989b Jun 08 '16 at 19:48
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@gt6989b In view of the (ordinals) tag, I guess $\omega$ is the lieast infinite ordinal. – bof Jun 08 '16 at 19:51
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@PedroSánchezTerraf Does the duplicate policy of math.se allow us to overlook the fact that $2\ne3$? – bof Jun 08 '16 at 19:52
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@bof Wait, $2\not=3$? Crud, there goes that paper. :P I think in this context it's close enough to be a duplicate. – Noah Schweber Jun 08 '16 at 19:53
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@bof both are examples of an $n$, and the reasoning is the same ;-) – Pedro Sánchez Terraf Jun 08 '16 at 19:53
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1Book recommendations: http://math.stackexchange.com/q/1491464/212120 – Pedro Sánchez Terraf Jun 08 '16 at 19:54
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1@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. – Brian M. Scott Jun 08 '16 at 20:34
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3One 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. – Brian M. Scott Jun 08 '16 at 20:38
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2@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 $2$. – Brian Tung Jun 08 '16 at 22:12
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What about Enderton? Is that any good? – Zzz Jun 11 '16 at 16:11
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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.
Pedro Sánchez Terraf
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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 $(\omega + 3) \cdot \omega = \omega \cdot \omega$.
Stefan Mesken
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I think you have to use normality (or continuity) of the ordinal product to justify last displayed step (i.e., $\sup { \omega \cdot (n+1) \mid n < \omega } = \omega \cdot \omega$.) – Pedro Sánchez Terraf Jun 09 '16 at 00:23
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@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. – Stefan Mesken Jun 09 '16 at 07:10
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That's perfect; the point I wanted to make is that you need to use a bit more than the bare definition of product. – Pedro Sánchez Terraf Jun 09 '16 at 11:07
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Could you please elaborate on how to get $\sup { (\omega + \omega) \cdot n \mid n < \omega } = \sup { \omega \cdot (n+1) \mid n < \omega }$? – Akira Dec 01 '18 at 08:18
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@LeAnhDung Note that, for $0 < n < \omega$, $\omega \cdot (n+1) = \omega \cdot n + \omega \le \omega \cdot n + \omega \cdot n = (\omega + \omega) \cdot n = (\omega \cdot 2) \cdot n = \omega \cdot (2n)$. – Stefan Mesken Dec 01 '18 at 11:02
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Please confirm my understanding! We have $(\omega + \omega) \cdot n=(\omega \cdot 2) \cdot n=\omega \cdot (2n)$. Thus our job is to prove $\sup { \omega \cdot (n+1) \mid n < \omega } = \sup { \omega \cdot (2n) \mid n < \omega }$. We have ${ \omega \cdot (2n) \mid 0<n < \omega } \subseteq { \omega \cdot (n+1) \mid 0<n < \omega }$ and $\forall 0<n < \omega,\exists 0<n' < \omega:\omega \cdot (n+1)<\omega \cdot (2n')$. The result is then followed. – Akira Dec 01 '18 at 11:35
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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*}
Eric Wofsey
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Jacob Wakem
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