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a) Give a direct proof of $\aleph_{\alpha} +\aleph_{\alpha} = \aleph_{\alpha}$ by expressing $\omega_{\alpha}$ as a disjoint union of two sets of cardinality $\aleph_{\alpha}$.

b) Give a direct proof of $n \cdot \aleph_{\alpha} = \aleph_{\alpha}$ by constructing a one-to-one mapping of $\omega_{\alpha}$ onto $n \times \omega_{\alpha}$, where $n$ is a positive natural number.

Remark: The exercises are taken from the book 'Introduction to Set Theory' by Hrbacek and Jech Chapter $7$ Section $2$

Attempt:

a) Note that cardinal addition $\kappa + \lambda $ is defined as $|A \cup B|$ where $|A|=\kappa$, $|B|=\lambda$ and $A \cap B=\emptyset$. Therefore, $\aleph_{\alpha} +\aleph_{\alpha} = \aleph_{\alpha}$ is the same as $|\omega_{\alpha} \cup \omega_{\alpha}|=|\omega_{\alpha}|$. But my problem is how to define a map by expressing $\omega_{\alpha}$ as a disjoint union of two sets of cardinality $\aleph_{\alpha}$ as $\omega_{\alpha}$ is not disjoint with itself.

b) Let $|C|=n$. I want to get a bijection $g:C \times \omega_{\alpha} \rightarrow \omega_{\alpha}$. My idea of the map is as follow: Arrange the pair $(1,1),(2,2),...,(n,n)$ as one column in a matrix. Then the second column is $(1,n+1),(2,n+2),...,(n,2n)$. So there will be a bijection between $C \times \omega_{\alpha}$ and $\omega_{\alpha}$. But I don't know how to formally define such function.

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You misunderstand the first part, $\aleph_\alpha+\aleph_\alpha$ is $|2\times\omega_\alpha|$ which is the disjoint union of $\{0\}\times\omega_\alpha$ and $\{1\}\times\omega_\alpha$. And to show that there is a bijection between that disjoint union and $\omega_\alpha$ you just need to notice that you can define a notion of "even" and "odd" ordinals, and then use the same proof (essentially) that $\Bbb N$ and $\Bbb Z$ have the same cardinality.

The second part follows almost trivially by induction from the first.

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  • $\begingroup$ So for part (a), assume that $A=\{ 0,1 \}$. Then I can define a map $f:A \times \omega_{\alpha} \rightarrow \omega_{\alpha}$ by $f(\{ 0 \}, \beta)=2 \cdot \beta$ and $f(\{ 1 \}, \beta)=2 \cdot \beta +1$ for all $\beta \in \omega_{\alpha}$? $\endgroup$ – Idonknow Oct 12 '14 at 5:59
  • $\begingroup$ No. Not quite. Because $2\cdot(\omega+1)=(2\cdot\omega)+1$ so the map you suggest is not injective. Try harder to think what a suitable definition for even and odd would be. $\endgroup$ – Asaf Karagila Oct 12 '14 at 6:04
  • $\begingroup$ Can give any hint? I try to modify the proof that $\mathbb{N}$ and $\mathbb{Z}$ have the same cardinality to get the mapping, but I got nowhere. $\endgroup$ – Idonknow Oct 12 '14 at 6:14
  • $\begingroup$ Did you define what it means for an ordinal to be even/odd? $\endgroup$ – Asaf Karagila Oct 12 '14 at 6:21
  • $\begingroup$ Nope. But I found it in wikipaedia. Now I am getting more and more confused. $\endgroup$ – Idonknow Oct 12 '14 at 6:24
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HINTS: For (b) note that every ordinal can be expressed uniquely in the form $\alpha+k$, where $\alpha$ is either $0$ or a limit ordinal, and $k\in\omega$. Think about reducing $k$ modulo $n$. If you can do that, you’ll get (a) easily as the special case $n=2$.

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  • $\begingroup$ Is my idea for (b) correct? $\endgroup$ – Idonknow Oct 12 '14 at 6:43
  • $\begingroup$ @Idonknow: It has a couple of problems. First, for $C$ you’re supposed to use the set $n=\{0,1,\ldots,n-1\}$. Much more important, however, is the fact that your matrix doesn’t actually represent $C\times\omega_\alpha$: if it did, the entries in the first row would be $\langle 1,0\rangle,\langle 1,1\rangle,\ldots,\langle 1,n-1\rangle$. $\endgroup$ – Brian M. Scott Oct 12 '14 at 6:47

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