A question about the proof about strongly inaccessible cardinal












0












$begingroup$


My textbook Introduction to Set Theory by Hrbacek and Jech presents Theorem 3.13 and its corresponding proof as follows:




enter image description here




Since the authors refer to Theorem 2.2, I post it here for reference:




enter image description here




My question concerns Part (b) of Theorem 3.13




Let $kappa$ be a strongly inaccessible cardinal. If each $Xin S$ has cardinality $< kappa$ and $|S| < kappa$, then $bigcup S$ has cardinality $< kappa$.



Proof:



Let $lambda = |S|$ and $mu = sup {|X| mid X in S}$. Then (by Theorem 2.2(a)) $mu < kappa$ because $kappa$ is regular, and $|bigcup S| le lambda cdot mu <kappa$.




In fact, I can not understand how the authors apply Theorem 2.2(a) to finish the proof. On the other hand, I have figured out another way to accomplish it as follows:




Let $lambda = |S|$ and $mu = sup {|X| mid X in S}$. We have $forall X in S:|X| < kappa implies mu le kappa$.



I claim that $mu < kappa$. If not, $mu = kappa$ and thus ${|X| mid X in S}$ is cofinal in $kappa$. It follows that $operatorname{cf}(kappa) le |{|X| mid X in S}| le |S|< kappa$ and thus $operatorname{cf}(kappa) < kappa$. Then $kappa$ is singular, which contradicts the fact that $kappa$ is regular.



Hence $mu < kappa$. We have $|bigcup S| le lambda cdot mu =max{lambda, mu} <kappa$.




Could you please explain how the authors apply Theorem 2.2(a) to finish the proof and verify my approach?



Thank you for your help!






elementary-set-theory proof-explanation cardinals







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    0












    $begingroup$


    My textbook Introduction to Set Theory by Hrbacek and Jech presents Theorem 3.13 and its corresponding proof as follows:




    enter image description here




    Since the authors refer to Theorem 2.2, I post it here for reference:




    enter image description here




    My question concerns Part (b) of Theorem 3.13




    Let $kappa$ be a strongly inaccessible cardinal. If each $Xin S$ has cardinality $< kappa$ and $|S| < kappa$, then $bigcup S$ has cardinality $< kappa$.



    Proof:



    Let $lambda = |S|$ and $mu = sup {|X| mid X in S}$. Then (by Theorem 2.2(a)) $mu < kappa$ because $kappa$ is regular, and $|bigcup S| le lambda cdot mu <kappa$.




    In fact, I can not understand how the authors apply Theorem 2.2(a) to finish the proof. On the other hand, I have figured out another way to accomplish it as follows:




    Let $lambda = |S|$ and $mu = sup {|X| mid X in S}$. We have $forall X in S:|X| < kappa implies mu le kappa$.



    I claim that $mu < kappa$. If not, $mu = kappa$ and thus ${|X| mid X in S}$ is cofinal in $kappa$. It follows that $operatorname{cf}(kappa) le |{|X| mid X in S}| le |S|< kappa$ and thus $operatorname{cf}(kappa) < kappa$. Then $kappa$ is singular, which contradicts the fact that $kappa$ is regular.



    Hence $mu < kappa$. We have $|bigcup S| le lambda cdot mu =max{lambda, mu} <kappa$.




    Could you please explain how the authors apply Theorem 2.2(a) to finish the proof and verify my approach?



    Thank you for your help!






    elementary-set-theory proof-explanation cardinals







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      My textbook Introduction to Set Theory by Hrbacek and Jech presents Theorem 3.13 and its corresponding proof as follows:




      enter image description here




      Since the authors refer to Theorem 2.2, I post it here for reference:




      enter image description here




      My question concerns Part (b) of Theorem 3.13




      Let $kappa$ be a strongly inaccessible cardinal. If each $Xin S$ has cardinality $< kappa$ and $|S| < kappa$, then $bigcup S$ has cardinality $< kappa$.



      Proof:



      Let $lambda = |S|$ and $mu = sup {|X| mid X in S}$. Then (by Theorem 2.2(a)) $mu < kappa$ because $kappa$ is regular, and $|bigcup S| le lambda cdot mu <kappa$.




      In fact, I can not understand how the authors apply Theorem 2.2(a) to finish the proof. On the other hand, I have figured out another way to accomplish it as follows:




      Let $lambda = |S|$ and $mu = sup {|X| mid X in S}$. We have $forall X in S:|X| < kappa implies mu le kappa$.



      I claim that $mu < kappa$. If not, $mu = kappa$ and thus ${|X| mid X in S}$ is cofinal in $kappa$. It follows that $operatorname{cf}(kappa) le |{|X| mid X in S}| le |S|< kappa$ and thus $operatorname{cf}(kappa) < kappa$. Then $kappa$ is singular, which contradicts the fact that $kappa$ is regular.



      Hence $mu < kappa$. We have $|bigcup S| le lambda cdot mu =max{lambda, mu} <kappa$.




      Could you please explain how the authors apply Theorem 2.2(a) to finish the proof and verify my approach?



      Thank you for your help!






      elementary-set-theory proof-explanation cardinals







      share|cite|improve this question











      $endgroup$




      My textbook Introduction to Set Theory by Hrbacek and Jech presents Theorem 3.13 and its corresponding proof as follows:




      enter image description here




      Since the authors refer to Theorem 2.2, I post it here for reference:




      enter image description here




      My question concerns Part (b) of Theorem 3.13




      Let $kappa$ be a strongly inaccessible cardinal. If each $Xin S$ has cardinality $< kappa$ and $|S| < kappa$, then $bigcup S$ has cardinality $< kappa$.



      Proof:



      Let $lambda = |S|$ and $mu = sup {|X| mid X in S}$. Then (by Theorem 2.2(a)) $mu < kappa$ because $kappa$ is regular, and $|bigcup S| le lambda cdot mu <kappa$.




      In fact, I can not understand how the authors apply Theorem 2.2(a) to finish the proof. On the other hand, I have figured out another way to accomplish it as follows:




      Let $lambda = |S|$ and $mu = sup {|X| mid X in S}$. We have $forall X in S:|X| < kappa implies mu le kappa$.



      I claim that $mu < kappa$. If not, $mu = kappa$ and thus ${|X| mid X in S}$ is cofinal in $kappa$. It follows that $operatorname{cf}(kappa) le |{|X| mid X in S}| le |S|< kappa$ and thus $operatorname{cf}(kappa) < kappa$. Then $kappa$ is singular, which contradicts the fact that $kappa$ is regular.



      Hence $mu < kappa$. We have $|bigcup S| le lambda cdot mu =max{lambda, mu} <kappa$.




      Could you please explain how the authors apply Theorem 2.2(a) to finish the proof and verify my approach?



      Thank you for your help!






      elementary-set-theory proof-explanation cardinals




      elementary-set-theory proof-explanation cardinals






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      edited Dec 30 '18 at 6:26









      Holo

      6,19721131




      6,19721131










      asked Dec 30 '18 at 6:08









      Le Anh DungLe Anh Dung

      1,2451621




      1,2451621






















          1 Answer
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          $begingroup$

          Your way is just the way they use the theorem, all of $Xin S$ are bounded by $gamma_X$, if $sup|X|$ is unbounded, the sequence generated from $gamma_X$ is unbounded, but the sequence is of length $lambda<kappa$, which is contradiction






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I seem to got it. Please check my reasoning! Assume the contrary that $mu = sup {|X| mid X in S} =kappa$ or equivalently ${|X| mid X in S}$ is unbounded in $kappa$. Then by every unbounded subset of $kappa$ has cardinality $kappa$ from Theorem 2.2(a), we have $|{|X| mid X in S}|=kappa$. On the other hand, $|{|X| mid X in S}| le |S|< kappa$. Thus $kappa < kappa$, which is a contradiction.
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:51










          • $begingroup$
            @LeAnhDung yes, correct
            $endgroup$
            – Holo
            Dec 30 '18 at 6:52










          • $begingroup$
            Thank you so much for your prompt reply!
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:53












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          $begingroup$

          Your way is just the way they use the theorem, all of $Xin S$ are bounded by $gamma_X$, if $sup|X|$ is unbounded, the sequence generated from $gamma_X$ is unbounded, but the sequence is of length $lambda<kappa$, which is contradiction






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I seem to got it. Please check my reasoning! Assume the contrary that $mu = sup {|X| mid X in S} =kappa$ or equivalently ${|X| mid X in S}$ is unbounded in $kappa$. Then by every unbounded subset of $kappa$ has cardinality $kappa$ from Theorem 2.2(a), we have $|{|X| mid X in S}|=kappa$. On the other hand, $|{|X| mid X in S}| le |S|< kappa$. Thus $kappa < kappa$, which is a contradiction.
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:51










          • $begingroup$
            @LeAnhDung yes, correct
            $endgroup$
            – Holo
            Dec 30 '18 at 6:52










          • $begingroup$
            Thank you so much for your prompt reply!
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:53
















          1












          $begingroup$

          Your way is just the way they use the theorem, all of $Xin S$ are bounded by $gamma_X$, if $sup|X|$ is unbounded, the sequence generated from $gamma_X$ is unbounded, but the sequence is of length $lambda<kappa$, which is contradiction






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I seem to got it. Please check my reasoning! Assume the contrary that $mu = sup {|X| mid X in S} =kappa$ or equivalently ${|X| mid X in S}$ is unbounded in $kappa$. Then by every unbounded subset of $kappa$ has cardinality $kappa$ from Theorem 2.2(a), we have $|{|X| mid X in S}|=kappa$. On the other hand, $|{|X| mid X in S}| le |S|< kappa$. Thus $kappa < kappa$, which is a contradiction.
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:51










          • $begingroup$
            @LeAnhDung yes, correct
            $endgroup$
            – Holo
            Dec 30 '18 at 6:52










          • $begingroup$
            Thank you so much for your prompt reply!
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:53














          1












          1








          1





          $begingroup$

          Your way is just the way they use the theorem, all of $Xin S$ are bounded by $gamma_X$, if $sup|X|$ is unbounded, the sequence generated from $gamma_X$ is unbounded, but the sequence is of length $lambda<kappa$, which is contradiction






          share|cite|improve this answer









          $endgroup$



          Your way is just the way they use the theorem, all of $Xin S$ are bounded by $gamma_X$, if $sup|X|$ is unbounded, the sequence generated from $gamma_X$ is unbounded, but the sequence is of length $lambda<kappa$, which is contradiction







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 6:24









          HoloHolo

          6,19721131




          6,19721131












          • $begingroup$
            I seem to got it. Please check my reasoning! Assume the contrary that $mu = sup {|X| mid X in S} =kappa$ or equivalently ${|X| mid X in S}$ is unbounded in $kappa$. Then by every unbounded subset of $kappa$ has cardinality $kappa$ from Theorem 2.2(a), we have $|{|X| mid X in S}|=kappa$. On the other hand, $|{|X| mid X in S}| le |S|< kappa$. Thus $kappa < kappa$, which is a contradiction.
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:51










          • $begingroup$
            @LeAnhDung yes, correct
            $endgroup$
            – Holo
            Dec 30 '18 at 6:52










          • $begingroup$
            Thank you so much for your prompt reply!
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:53


















          • $begingroup$
            I seem to got it. Please check my reasoning! Assume the contrary that $mu = sup {|X| mid X in S} =kappa$ or equivalently ${|X| mid X in S}$ is unbounded in $kappa$. Then by every unbounded subset of $kappa$ has cardinality $kappa$ from Theorem 2.2(a), we have $|{|X| mid X in S}|=kappa$. On the other hand, $|{|X| mid X in S}| le |S|< kappa$. Thus $kappa < kappa$, which is a contradiction.
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:51










          • $begingroup$
            @LeAnhDung yes, correct
            $endgroup$
            – Holo
            Dec 30 '18 at 6:52










          • $begingroup$
            Thank you so much for your prompt reply!
            $endgroup$
            – Le Anh Dung
            Dec 30 '18 at 6:53
















          $begingroup$
          I seem to got it. Please check my reasoning! Assume the contrary that $mu = sup {|X| mid X in S} =kappa$ or equivalently ${|X| mid X in S}$ is unbounded in $kappa$. Then by every unbounded subset of $kappa$ has cardinality $kappa$ from Theorem 2.2(a), we have $|{|X| mid X in S}|=kappa$. On the other hand, $|{|X| mid X in S}| le |S|< kappa$. Thus $kappa < kappa$, which is a contradiction.
          $endgroup$
          – Le Anh Dung
          Dec 30 '18 at 6:51




          $begingroup$
          I seem to got it. Please check my reasoning! Assume the contrary that $mu = sup {|X| mid X in S} =kappa$ or equivalently ${|X| mid X in S}$ is unbounded in $kappa$. Then by every unbounded subset of $kappa$ has cardinality $kappa$ from Theorem 2.2(a), we have $|{|X| mid X in S}|=kappa$. On the other hand, $|{|X| mid X in S}| le |S|< kappa$. Thus $kappa < kappa$, which is a contradiction.
          $endgroup$
          – Le Anh Dung
          Dec 30 '18 at 6:51












          $begingroup$
          @LeAnhDung yes, correct
          $endgroup$
          – Holo
          Dec 30 '18 at 6:52




          $begingroup$
          @LeAnhDung yes, correct
          $endgroup$
          – Holo
          Dec 30 '18 at 6:52












          $begingroup$
          Thank you so much for your prompt reply!
          $endgroup$
          – Le Anh Dung
          Dec 30 '18 at 6:53




          $begingroup$
          Thank you so much for your prompt reply!
          $endgroup$
          – Le Anh Dung
          Dec 30 '18 at 6:53


















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