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Q1

Prove that the set $\mathbb{R}^+$ of the positive reals can be written as the union of two non-empty sets, say $A$ & $B$ , both these set are closed unnder addition.

Q2

$\aleph_\omega ,\aleph_{\omega_1}$ and $\aleph_{\omega_2}$ It is true that these three are smallest singular cardinals (by order, I mean the $1^{st}$, $2^{nd}$ and $3^{rd}$).

Michael Hardy
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Alexandro Mrose
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1 Answers1

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The first answer is the same usual use of Zorn's lemma. Simply find the right partial order (hint, pairs of disjoint sets closed under addition).

The full details appear in Using Zorn's lemma show that $\mathbb R^+$ is the disjoint union of two sets closed under addition.

The second question is plain false. Recall that $\aleph_\alpha$ is singular when $\alpha$ is a limit ordinal and $\alpha<\omega_\alpha$ (it is possible to have singular cardinal with equality, though). Therefore the first three singular cardinals would be those whose indices are the first three limit ordinals.

Remember that there are $\aleph_1$ limit ordinals smaller than $\omega_1$, so it cannot possibly be the second singular cardinal.

Community
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Asaf Karagila
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  • Thanks for your reply. But I have seen that $\aleph_\omega$ is the smallest cardinal. See this [link](books.google.co.uk/books?id=RBrWwKVbmMUC&pg=PA526&lpg=PA526&dq="smallest+singular+cardinal"&source=bl&ots=msq21otttU&sig=9hxaY4UpKzaUdyP2aQgbUr8j_Sw&hl=en&sa=X&ei=T7dDUbezDfSW0QXbjoGICA&ved=0CEMQ6AEwBA#v=onepage&q="smallest singular cardinal"&f=false) – Alexandro Mrose Mar 16 '13 at 00:08
  • @AlexandroMrose: How does that contradict what I said? – Asaf Karagila Mar 16 '13 at 00:08
  • [books.google.co.uk/books?id=RBrWwKVbmMUC&pg=PA526&lpg=PA526&dq="smallest+singular+cardinal"&source=bl&ots=msq21otttU&sig=9hxaY4UpKzaUdyP2aQgbUr8j_Sw&hl=en&sa=X&ei=T7dDUbezDfSW0QXbjoGICA&ved=0CEMQ6AEwBA#v=onepage&q="smallest singular cardinal"&f=false] (Check this link) – Alexandro Mrose Mar 16 '13 at 00:15
  • Either you haven't read my answer, or you haven't read my comment. How does this contradict my answer? – Asaf Karagila Mar 16 '13 at 00:17
  • I have read your answer, but I do not know much about set theory. I do not understand well, I only want to know which are the 1st, 2nd and 3rd sigular cardinals. Thanks in advance – Alexandro Mrose Mar 16 '13 at 00:23
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    Alexandro, $\omega$ is the least limit ordinal so $\aleph_\omega$ is indeed the least singular cardinal. I'm not so clear as for why you are asking this if you don't know enough about set theory. – Asaf Karagila Mar 16 '13 at 00:28
  • Dear Asaf, by this clear reply I have understood your previous comments as well. I will take it in the next semester, so I need at least some idea. I really thank you. Cheers – Alexandro Mrose Mar 16 '13 at 00:35
  • Alexandro, then you may want to read more about ordinals first. Then about cardinals. The answer to your question would become quite apparent (if not immediately, then by applying my hint above). As for the first question, using Zorn's lemma is a standard argument, although it can get quite laborious. If you have experience using Zorn's lemma you may find my linked answer (hover over the gray area) helpful, and if you don't have experience using Zorn's lemma you may want to start with standard uses such as existence of a basis of a vector space, and so on. – Asaf Karagila Mar 16 '13 at 00:41
  • Asaf, Hum! this is a very good comment I liked it. Again thanks for these valuable words. – Alexandro Mrose Mar 16 '13 at 00:46