Is $aleph_1= aleph_0+1 $ wrong?

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My understanding is that any cardinality is always an integer because it expresses how many elements are in a given set. And I read that $aleph_1$ is the next smallest cardinality that's larger than $aleph_0$, so it seems to me that $aleph_1 = aleph_0+1$. However I don't see this equation anywhere on the Internet so I guess I'm wrong. Where am I wrong?




elementary-set-theory cardinals




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  • 2




    Only the finite cardinalities are (nonnegative) integers; the rest are, well, cardinals, which are different. $aleph_0 + 1$ is just $aleph_0$ again.
    – Qiaochu Yuan
    Jul 23 at 7:18











  • What is true is that $aleph_1=aleph_0+1$.
    – Lord Shark the Unknown
    Jul 23 at 7:19










  • $aleph_0+1=aleph_1gtaleph_0=aleph_0+1.$
    – bof
    Jul 23 at 7:19










  • @QiaochuYuan Is there a special symbol or something to express the value of $aleph_1$ - $aleph_0$?
    – stacko
    Jul 23 at 7:21






  • 1




    @stacko this is independent of $ZFC$(see en.wikipedia.org/wiki/Continuum_hypothesis)
    – Holo
    Jul 23 at 7:34















up vote
0
down vote

favorite












My understanding is that any cardinality is always an integer because it expresses how many elements are in a given set. And I read that $aleph_1$ is the next smallest cardinality that's larger than $aleph_0$, so it seems to me that $aleph_1 = aleph_0+1$. However I don't see this equation anywhere on the Internet so I guess I'm wrong. Where am I wrong?




elementary-set-theory cardinals




share|cite|improve this question

















  • 2




    Only the finite cardinalities are (nonnegative) integers; the rest are, well, cardinals, which are different. $aleph_0 + 1$ is just $aleph_0$ again.
    – Qiaochu Yuan
    Jul 23 at 7:18











  • What is true is that $aleph_1=aleph_0+1$.
    – Lord Shark the Unknown
    Jul 23 at 7:19










  • $aleph_0+1=aleph_1gtaleph_0=aleph_0+1.$
    – bof
    Jul 23 at 7:19










  • @QiaochuYuan Is there a special symbol or something to express the value of $aleph_1$ - $aleph_0$?
    – stacko
    Jul 23 at 7:21






  • 1




    @stacko this is independent of $ZFC$(see en.wikipedia.org/wiki/Continuum_hypothesis)
    – Holo
    Jul 23 at 7:34













up vote
0
down vote

favorite









up vote
0
down vote

favorite











My understanding is that any cardinality is always an integer because it expresses how many elements are in a given set. And I read that $aleph_1$ is the next smallest cardinality that's larger than $aleph_0$, so it seems to me that $aleph_1 = aleph_0+1$. However I don't see this equation anywhere on the Internet so I guess I'm wrong. Where am I wrong?




elementary-set-theory cardinals




share|cite|improve this question













My understanding is that any cardinality is always an integer because it expresses how many elements are in a given set. And I read that $aleph_1$ is the next smallest cardinality that's larger than $aleph_0$, so it seems to me that $aleph_1 = aleph_0+1$. However I don't see this equation anywhere on the Internet so I guess I'm wrong. Where am I wrong?





elementary-set-theory cardinals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 23 at 7:20









Lord Shark the Unknown

85.2k950111




85.2k950111









asked Jul 23 at 7:15









stacko

15414




15414







  • 2




    Only the finite cardinalities are (nonnegative) integers; the rest are, well, cardinals, which are different. $aleph_0 + 1$ is just $aleph_0$ again.
    – Qiaochu Yuan
    Jul 23 at 7:18











  • What is true is that $aleph_1=aleph_0+1$.
    – Lord Shark the Unknown
    Jul 23 at 7:19










  • $aleph_0+1=aleph_1gtaleph_0=aleph_0+1.$
    – bof
    Jul 23 at 7:19










  • @QiaochuYuan Is there a special symbol or something to express the value of $aleph_1$ - $aleph_0$?
    – stacko
    Jul 23 at 7:21






  • 1




    @stacko this is independent of $ZFC$(see en.wikipedia.org/wiki/Continuum_hypothesis)
    – Holo
    Jul 23 at 7:34













  • 2




    Only the finite cardinalities are (nonnegative) integers; the rest are, well, cardinals, which are different. $aleph_0 + 1$ is just $aleph_0$ again.
    – Qiaochu Yuan
    Jul 23 at 7:18











  • What is true is that $aleph_1=aleph_0+1$.
    – Lord Shark the Unknown
    Jul 23 at 7:19










  • $aleph_0+1=aleph_1gtaleph_0=aleph_0+1.$
    – bof
    Jul 23 at 7:19










  • @QiaochuYuan Is there a special symbol or something to express the value of $aleph_1$ - $aleph_0$?
    – stacko
    Jul 23 at 7:21






  • 1




    @stacko this is independent of $ZFC$(see en.wikipedia.org/wiki/Continuum_hypothesis)
    – Holo
    Jul 23 at 7:34








2




2




Only the finite cardinalities are (nonnegative) integers; the rest are, well, cardinals, which are different. $aleph_0 + 1$ is just $aleph_0$ again.
– Qiaochu Yuan
Jul 23 at 7:18





Only the finite cardinalities are (nonnegative) integers; the rest are, well, cardinals, which are different. $aleph_0 + 1$ is just $aleph_0$ again.
– Qiaochu Yuan
Jul 23 at 7:18













What is true is that $aleph_1=aleph_0+1$.
– Lord Shark the Unknown
Jul 23 at 7:19




What is true is that $aleph_1=aleph_0+1$.
– Lord Shark the Unknown
Jul 23 at 7:19












$aleph_0+1=aleph_1gtaleph_0=aleph_0+1.$
– bof
Jul 23 at 7:19




$aleph_0+1=aleph_1gtaleph_0=aleph_0+1.$
– bof
Jul 23 at 7:19












@QiaochuYuan Is there a special symbol or something to express the value of $aleph_1$ - $aleph_0$?
– stacko
Jul 23 at 7:21




@QiaochuYuan Is there a special symbol or something to express the value of $aleph_1$ - $aleph_0$?
– stacko
Jul 23 at 7:21




1




1




@stacko this is independent of $ZFC$(see en.wikipedia.org/wiki/Continuum_hypothesis)
– Holo
Jul 23 at 7:34





@stacko this is independent of $ZFC$(see en.wikipedia.org/wiki/Continuum_hypothesis)
– Holo
Jul 23 at 7:34











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The wrong part is that you took what you know about finite cardinals and extend it to infinite ones.



Assuming choice $aleph_0$ is the smallest cardinal that is greater than all the finite cardinal, the first limit ordinal. To advance to $aleph_1$ we need to find a set such that there is no injective from $A$ to $Bbb N$, but $aleph_0+1=|Bbb Ncup0|=|Bbb N|=aleph_0$.



Even more: for $kappa,nu$ cardinals such that one of them is infinite we have $kappa+nu=kappacdotnu=maxkappa,nu$






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  • @Arthur yes, but it is not the smallest without it
    – Holo
    Jul 23 at 7:30






  • 1




    We only need countable choice for $aleph_0$ to be the smallest infinite ordinal (and I think it's minimal regardless), but fair enough. We do need choice for $max$ to be well-defined, though.
    – Arthur
    Jul 23 at 7:36











  • @Arthur indeed even without choice every infinite cardinal is either not comparable or greater equal to $aleph_0$
    – Holo
    Jul 23 at 7:59










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up vote
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The wrong part is that you took what you know about finite cardinals and extend it to infinite ones.



Assuming choice $aleph_0$ is the smallest cardinal that is greater than all the finite cardinal, the first limit ordinal. To advance to $aleph_1$ we need to find a set such that there is no injective from $A$ to $Bbb N$, but $aleph_0+1=|Bbb Ncup0|=|Bbb N|=aleph_0$.



Even more: for $kappa,nu$ cardinals such that one of them is infinite we have $kappa+nu=kappacdotnu=maxkappa,nu$






share|cite|improve this answer





















  • @Arthur yes, but it is not the smallest without it
    – Holo
    Jul 23 at 7:30






  • 1




    We only need countable choice for $aleph_0$ to be the smallest infinite ordinal (and I think it's minimal regardless), but fair enough. We do need choice for $max$ to be well-defined, though.
    – Arthur
    Jul 23 at 7:36











  • @Arthur indeed even without choice every infinite cardinal is either not comparable or greater equal to $aleph_0$
    – Holo
    Jul 23 at 7:59














up vote
0
down vote













The wrong part is that you took what you know about finite cardinals and extend it to infinite ones.



Assuming choice $aleph_0$ is the smallest cardinal that is greater than all the finite cardinal, the first limit ordinal. To advance to $aleph_1$ we need to find a set such that there is no injective from $A$ to $Bbb N$, but $aleph_0+1=|Bbb Ncup0|=|Bbb N|=aleph_0$.



Even more: for $kappa,nu$ cardinals such that one of them is infinite we have $kappa+nu=kappacdotnu=maxkappa,nu$






share|cite|improve this answer





















  • @Arthur yes, but it is not the smallest without it
    – Holo
    Jul 23 at 7:30






  • 1




    We only need countable choice for $aleph_0$ to be the smallest infinite ordinal (and I think it's minimal regardless), but fair enough. We do need choice for $max$ to be well-defined, though.
    – Arthur
    Jul 23 at 7:36











  • @Arthur indeed even without choice every infinite cardinal is either not comparable or greater equal to $aleph_0$
    – Holo
    Jul 23 at 7:59












up vote
0
down vote










up vote
0
down vote









The wrong part is that you took what you know about finite cardinals and extend it to infinite ones.



Assuming choice $aleph_0$ is the smallest cardinal that is greater than all the finite cardinal, the first limit ordinal. To advance to $aleph_1$ we need to find a set such that there is no injective from $A$ to $Bbb N$, but $aleph_0+1=|Bbb Ncup0|=|Bbb N|=aleph_0$.



Even more: for $kappa,nu$ cardinals such that one of them is infinite we have $kappa+nu=kappacdotnu=maxkappa,nu$






share|cite|improve this answer













The wrong part is that you took what you know about finite cardinals and extend it to infinite ones.



Assuming choice $aleph_0$ is the smallest cardinal that is greater than all the finite cardinal, the first limit ordinal. To advance to $aleph_1$ we need to find a set such that there is no injective from $A$ to $Bbb N$, but $aleph_0+1=|Bbb Ncup0|=|Bbb N|=aleph_0$.



Even more: for $kappa,nu$ cardinals such that one of them is infinite we have $kappa+nu=kappacdotnu=maxkappa,nu$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 23 at 7:25









Holo

4,1972528




4,1972528











  • @Arthur yes, but it is not the smallest without it
    – Holo
    Jul 23 at 7:30






  • 1




    We only need countable choice for $aleph_0$ to be the smallest infinite ordinal (and I think it's minimal regardless), but fair enough. We do need choice for $max$ to be well-defined, though.
    – Arthur
    Jul 23 at 7:36











  • @Arthur indeed even without choice every infinite cardinal is either not comparable or greater equal to $aleph_0$
    – Holo
    Jul 23 at 7:59
















  • @Arthur yes, but it is not the smallest without it
    – Holo
    Jul 23 at 7:30






  • 1




    We only need countable choice for $aleph_0$ to be the smallest infinite ordinal (and I think it's minimal regardless), but fair enough. We do need choice for $max$ to be well-defined, though.
    – Arthur
    Jul 23 at 7:36











  • @Arthur indeed even without choice every infinite cardinal is either not comparable or greater equal to $aleph_0$
    – Holo
    Jul 23 at 7:59















@Arthur yes, but it is not the smallest without it
– Holo
Jul 23 at 7:30




@Arthur yes, but it is not the smallest without it
– Holo
Jul 23 at 7:30




1




1




We only need countable choice for $aleph_0$ to be the smallest infinite ordinal (and I think it's minimal regardless), but fair enough. We do need choice for $max$ to be well-defined, though.
– Arthur
Jul 23 at 7:36





We only need countable choice for $aleph_0$ to be the smallest infinite ordinal (and I think it's minimal regardless), but fair enough. We do need choice for $max$ to be well-defined, though.
– Arthur
Jul 23 at 7:36













@Arthur indeed even without choice every infinite cardinal is either not comparable or greater equal to $aleph_0$
– Holo
Jul 23 at 7:59




@Arthur indeed even without choice every infinite cardinal is either not comparable or greater equal to $aleph_0$
– Holo
Jul 23 at 7:59












 

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