Equivalent condition for a cardinal number to be of cofinality $aleph_0$












0












$begingroup$


Let cf$(alpha)$ denotes the co-final of the transfinite cardinal number $alpha.$
For every successor cardinal $alpha$ we have cf$(alpha)=alphaneq aleph_0.$
Thus cf$(alpha) = aleph_0,$ implies $alpha$ is a limit cardinal.



We can find some limit cardinal number $alpha,$ for example $alpha=aleph_{omega}$ with cf$(alpha)=aleph_0.$ Now, can we fined all (limit) cardinal numbers which satisfies
cf$(alpha)=aleph_0?$






set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    $aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:12










  • $begingroup$
    Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:14












  • $begingroup$
    OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
    $endgroup$
    – Ali Bayati
    Jan 9 at 8:27






  • 2




    $begingroup$
    I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:48
















0












$begingroup$


Let cf$(alpha)$ denotes the co-final of the transfinite cardinal number $alpha.$
For every successor cardinal $alpha$ we have cf$(alpha)=alphaneq aleph_0.$
Thus cf$(alpha) = aleph_0,$ implies $alpha$ is a limit cardinal.



We can find some limit cardinal number $alpha,$ for example $alpha=aleph_{omega}$ with cf$(alpha)=aleph_0.$ Now, can we fined all (limit) cardinal numbers which satisfies
cf$(alpha)=aleph_0?$






set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    $aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:12










  • $begingroup$
    Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:14












  • $begingroup$
    OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
    $endgroup$
    – Ali Bayati
    Jan 9 at 8:27






  • 2




    $begingroup$
    I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:48














0












0








0





$begingroup$


Let cf$(alpha)$ denotes the co-final of the transfinite cardinal number $alpha.$
For every successor cardinal $alpha$ we have cf$(alpha)=alphaneq aleph_0.$
Thus cf$(alpha) = aleph_0,$ implies $alpha$ is a limit cardinal.



We can find some limit cardinal number $alpha,$ for example $alpha=aleph_{omega}$ with cf$(alpha)=aleph_0.$ Now, can we fined all (limit) cardinal numbers which satisfies
cf$(alpha)=aleph_0?$






set-theory







share|cite|improve this question











$endgroup$




Let cf$(alpha)$ denotes the co-final of the transfinite cardinal number $alpha.$
For every successor cardinal $alpha$ we have cf$(alpha)=alphaneq aleph_0.$
Thus cf$(alpha) = aleph_0,$ implies $alpha$ is a limit cardinal.



We can find some limit cardinal number $alpha,$ for example $alpha=aleph_{omega}$ with cf$(alpha)=aleph_0.$ Now, can we fined all (limit) cardinal numbers which satisfies
cf$(alpha)=aleph_0?$






set-theory




set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 8:22







Ali Bayati

















asked Jan 9 at 8:11









Ali BayatiAli Bayati

344




344












  • $begingroup$
    $aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:12










  • $begingroup$
    Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:14












  • $begingroup$
    OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
    $endgroup$
    – Ali Bayati
    Jan 9 at 8:27






  • 2




    $begingroup$
    I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:48


















  • $begingroup$
    $aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:12










  • $begingroup$
    Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:14












  • $begingroup$
    OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
    $endgroup$
    – Ali Bayati
    Jan 9 at 8:27






  • 2




    $begingroup$
    I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
    $endgroup$
    – Asaf Karagila♦
    Jan 9 at 8:48
















$begingroup$
$aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:12




$begingroup$
$aleph_omega$ has cofinality $aleph_0$, but $aleph_{omega_1}$ has cofinality $aleph_1$. At least assuming the axiom of choice.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:12












$begingroup$
Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:14






$begingroup$
Prove that $operatorname{cf}(aleph_alpha)=operatorname{cf}(alpha)$ for limit ordinals (although cofinality of arbitrary ordinals is easier to measure with order types rather than cardinals, but for initial ordinals it is the same).
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:14














$begingroup$
OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
$endgroup$
– Ali Bayati
Jan 9 at 8:27




$begingroup$
OK! I wrote wrong. But, is there an equivalent condition for cf$(alpha)=aleph_0?$
$endgroup$
– Ali Bayati
Jan 9 at 8:27




2




2




$begingroup$
I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:48




$begingroup$
I don't understand your last two comments. At all. My second comment provides an equivalent definition. It is not predicative, since you can have $aleph_alpha=alpha$ with cofinality $aleph_0$.
$endgroup$
– Asaf Karagila♦
Jan 9 at 8:48










1 Answer
1






active

oldest

votes


















1












$begingroup$

Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.



Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:




$aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.




As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).



So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.



Proving the fact above is a good exercise; here are a couple hints:




  • In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?


  • In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?







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    $begingroup$

    Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.



    Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:




    $aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.




    As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).



    So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.



    Proving the fact above is a good exercise; here are a couple hints:




    • In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?


    • In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?







    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.



      Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:




      $aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.




      As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).



      So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.



      Proving the fact above is a good exercise; here are a couple hints:




      • In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?


      • In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?







      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.



        Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:




        $aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.




        As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).



        So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.



        Proving the fact above is a good exercise; here are a couple hints:




        • In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?


        • In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?







        share|cite|improve this answer









        $endgroup$



        Asaf Karagila has essentially answered this question in comments (and if he posts an answer I'll delete this one for precedence reasons), but let me flesh it out a bit.



        Remember that cofinality makes sense for arbitrary infinite limit ordinals, not just infinite cardinals. The characterization is:




        $aleph_alpha$ has cofinality $omega$ iff $alpha=0$ or $alpha$ (is a limit ordinal and) has cofinality $omega$.




        As an aside, remember that $aleph_alpha$ and $omega_alpha$ are literally the same thing, the difference between the notation is merely that the former indicates that we're thinking about only cardinals while the second indicates that we're thinking about general ordinals. I'm going to use both notations since in this particular question I think it will be clearer, but really I think the ideal approach is to ditch the $aleph$-notation completely except for situations when we're using cardinal arithmetic - that is cardinal addition, multiplication, and exponentiation (which we're not doing here).



        So, for example, the cardinals $aleph_omega, aleph_{omega^3+omega^2cdot 13}$, and $aleph_{omega_{omega_omega}}$ each have cofinality $omega$, while $aleph_{37}$ and $aleph_{omega_1}$ do not since $cf(37)=1$ and $cf(omega_1)=omega_1$.



        Proving the fact above is a good exercise; here are a couple hints:




        • In the right-to-left direction, suppose $cf(alpha)=omega$. Let $(beta_i)_{iinomega}$ be cofinal in $alpha$; what can you say about the sequence $(aleph_{beta_i})_{iinomega}$?


        • In the left-to-right direction, we do the same thing but in the other direction. Suppose $aleph_alpha$ has cofinality $omega$ and $alphanot=0$. Let $(aleph_{beta_i})_{iinomega}$ be cofinal in $aleph_alpha$: what can you say about the sequence $(beta_i)_{iinomega}$, and what does that tell us about $alpha$?








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 20 at 14:09









        Noah SchweberNoah Schweber

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