# $(\omega +3)\cdot\omega=\omega\cdot\omega$ [duplicate]

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?

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

• What is $\omega$ here? – gt6989b Jun 8 '16 at 19:48
• Book recommendations: math.stackexchange.com/q/1491464/212120 – Pedro Sánchez Terraf Jun 8 '16 at 19:54
• @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 8 '16 at 20:34
• 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. – Brian M. Scott Jun 8 '16 at 20:38
• @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 8 '16 at 22:12

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

• 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 9 '16 at 0:23
• @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 9 '16 at 7:10
• 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 9 '16 at 11:07
• @LeAnhDung Yep, that's it. – Stefan Mesken Dec 1 '18 at 11:37
• @LeAnhDung You're most welcome! – Stefan Mesken Dec 1 '18 at 11:38

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.

\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*}

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