Questions about Aleph-Aleph-Null












2














Note: I apologize in advance for not using proper notation on some of these values, but this is literally my first post on this site and I do not know how to display these values correctly.



I recently was looking up facts about different cardinalities of infinity for a book idea, when I found a post made several years ago about ℵ(ℵ(0))



If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?



In this post people are talk about the difference between cardinal numbers and how ℵℵ0 should instead be ℵ(ω). The responses to the post then go on to talk about ℵ(ω+1), ℵ(ω+2), and so on.



Anyways, my understanding of the different values of ℵ was that they corresponded to the cardinalities of infinite sets, with ℵ(0) being the cardinality of the set of all natural numbers, and that if set X has cardinality of ℵ(a), then the cardinality of the powerset of X would be N(a+1).



With this in mind, I always imagined that if a set Y had cardinality ℵ(0), and you found its powerset, and then you found the powerset of that set, and then you found the powerset of THAT set, and repeated the process infinitely you would get a set with cardinality ℵ(ℵ(0)).



So, I guess my question is, in the discussion linked above, when people are talking about ℵ(ω+1), how is that possible? Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?



I hope I worded this question in a way that people will understand, and thanks in advance into any insight you can give me about all this.










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  • For the math symbols: math.meta.stackexchange.com/questions/5020/…
    – Holo
    1 hour ago






  • 1




    Here is maybe a better way to think of it: $aleph_omega$ is the smallest cardinal which is strictly greater than $aleph_n$ for all $n$. Then, it's perfectly reasonable to say that $2^{aleph_omega}$ is bigger than $aleph_omega$.
    – pseudocydonia
    1 hour ago






  • 1




    However, in general you can't assume that the power set of $aleph_a$ is equal to $aleph_{a+1}$ - that amounts to asserting the full version of the Generalized Continuum Hypothesis!
    – pseudocydonia
    1 hour ago
















2














Note: I apologize in advance for not using proper notation on some of these values, but this is literally my first post on this site and I do not know how to display these values correctly.



I recently was looking up facts about different cardinalities of infinity for a book idea, when I found a post made several years ago about ℵ(ℵ(0))



If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?



In this post people are talk about the difference between cardinal numbers and how ℵℵ0 should instead be ℵ(ω). The responses to the post then go on to talk about ℵ(ω+1), ℵ(ω+2), and so on.



Anyways, my understanding of the different values of ℵ was that they corresponded to the cardinalities of infinite sets, with ℵ(0) being the cardinality of the set of all natural numbers, and that if set X has cardinality of ℵ(a), then the cardinality of the powerset of X would be N(a+1).



With this in mind, I always imagined that if a set Y had cardinality ℵ(0), and you found its powerset, and then you found the powerset of that set, and then you found the powerset of THAT set, and repeated the process infinitely you would get a set with cardinality ℵ(ℵ(0)).



So, I guess my question is, in the discussion linked above, when people are talking about ℵ(ω+1), how is that possible? Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?



I hope I worded this question in a way that people will understand, and thanks in advance into any insight you can give me about all this.










share|cite|improve this question







New contributor




Patrick Malone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • For the math symbols: math.meta.stackexchange.com/questions/5020/…
    – Holo
    1 hour ago






  • 1




    Here is maybe a better way to think of it: $aleph_omega$ is the smallest cardinal which is strictly greater than $aleph_n$ for all $n$. Then, it's perfectly reasonable to say that $2^{aleph_omega}$ is bigger than $aleph_omega$.
    – pseudocydonia
    1 hour ago






  • 1




    However, in general you can't assume that the power set of $aleph_a$ is equal to $aleph_{a+1}$ - that amounts to asserting the full version of the Generalized Continuum Hypothesis!
    – pseudocydonia
    1 hour ago














2












2








2







Note: I apologize in advance for not using proper notation on some of these values, but this is literally my first post on this site and I do not know how to display these values correctly.



I recently was looking up facts about different cardinalities of infinity for a book idea, when I found a post made several years ago about ℵ(ℵ(0))



If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?



In this post people are talk about the difference between cardinal numbers and how ℵℵ0 should instead be ℵ(ω). The responses to the post then go on to talk about ℵ(ω+1), ℵ(ω+2), and so on.



Anyways, my understanding of the different values of ℵ was that they corresponded to the cardinalities of infinite sets, with ℵ(0) being the cardinality of the set of all natural numbers, and that if set X has cardinality of ℵ(a), then the cardinality of the powerset of X would be N(a+1).



With this in mind, I always imagined that if a set Y had cardinality ℵ(0), and you found its powerset, and then you found the powerset of that set, and then you found the powerset of THAT set, and repeated the process infinitely you would get a set with cardinality ℵ(ℵ(0)).



So, I guess my question is, in the discussion linked above, when people are talking about ℵ(ω+1), how is that possible? Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?



I hope I worded this question in a way that people will understand, and thanks in advance into any insight you can give me about all this.










share|cite|improve this question







New contributor




Patrick Malone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Note: I apologize in advance for not using proper notation on some of these values, but this is literally my first post on this site and I do not know how to display these values correctly.



I recently was looking up facts about different cardinalities of infinity for a book idea, when I found a post made several years ago about ℵ(ℵ(0))



If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?



In this post people are talk about the difference between cardinal numbers and how ℵℵ0 should instead be ℵ(ω). The responses to the post then go on to talk about ℵ(ω+1), ℵ(ω+2), and so on.



Anyways, my understanding of the different values of ℵ was that they corresponded to the cardinalities of infinite sets, with ℵ(0) being the cardinality of the set of all natural numbers, and that if set X has cardinality of ℵ(a), then the cardinality of the powerset of X would be N(a+1).



With this in mind, I always imagined that if a set Y had cardinality ℵ(0), and you found its powerset, and then you found the powerset of that set, and then you found the powerset of THAT set, and repeated the process infinitely you would get a set with cardinality ℵ(ℵ(0)).



So, I guess my question is, in the discussion linked above, when people are talking about ℵ(ω+1), how is that possible? Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?



I hope I worded this question in a way that people will understand, and thanks in advance into any insight you can give me about all this.







elementary-set-theory cardinals






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Patrick Malone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Patrick Malone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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asked 1 hour ago









Patrick Malone

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111




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New contributor





Patrick Malone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Patrick Malone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • For the math symbols: math.meta.stackexchange.com/questions/5020/…
    – Holo
    1 hour ago






  • 1




    Here is maybe a better way to think of it: $aleph_omega$ is the smallest cardinal which is strictly greater than $aleph_n$ for all $n$. Then, it's perfectly reasonable to say that $2^{aleph_omega}$ is bigger than $aleph_omega$.
    – pseudocydonia
    1 hour ago






  • 1




    However, in general you can't assume that the power set of $aleph_a$ is equal to $aleph_{a+1}$ - that amounts to asserting the full version of the Generalized Continuum Hypothesis!
    – pseudocydonia
    1 hour ago


















  • For the math symbols: math.meta.stackexchange.com/questions/5020/…
    – Holo
    1 hour ago






  • 1




    Here is maybe a better way to think of it: $aleph_omega$ is the smallest cardinal which is strictly greater than $aleph_n$ for all $n$. Then, it's perfectly reasonable to say that $2^{aleph_omega}$ is bigger than $aleph_omega$.
    – pseudocydonia
    1 hour ago






  • 1




    However, in general you can't assume that the power set of $aleph_a$ is equal to $aleph_{a+1}$ - that amounts to asserting the full version of the Generalized Continuum Hypothesis!
    – pseudocydonia
    1 hour ago
















For the math symbols: math.meta.stackexchange.com/questions/5020/…
– Holo
1 hour ago




For the math symbols: math.meta.stackexchange.com/questions/5020/…
– Holo
1 hour ago




1




1




Here is maybe a better way to think of it: $aleph_omega$ is the smallest cardinal which is strictly greater than $aleph_n$ for all $n$. Then, it's perfectly reasonable to say that $2^{aleph_omega}$ is bigger than $aleph_omega$.
– pseudocydonia
1 hour ago




Here is maybe a better way to think of it: $aleph_omega$ is the smallest cardinal which is strictly greater than $aleph_n$ for all $n$. Then, it's perfectly reasonable to say that $2^{aleph_omega}$ is bigger than $aleph_omega$.
– pseudocydonia
1 hour ago




1




1




However, in general you can't assume that the power set of $aleph_a$ is equal to $aleph_{a+1}$ - that amounts to asserting the full version of the Generalized Continuum Hypothesis!
– pseudocydonia
1 hour ago




However, in general you can't assume that the power set of $aleph_a$ is equal to $aleph_{a+1}$ - that amounts to asserting the full version of the Generalized Continuum Hypothesis!
– pseudocydonia
1 hour ago










2 Answers
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6














The reason it should really be $aleph_omega$ instead of $aleph_{aleph_0}$ is that we think of $aleph_alpha$ for ordinals $alpha$. Yes, $aleph_0=omega$, but we write $omega$ when we think of it as an ordinal instead of a cardinal.



The rest of your question is based on a fundamental misunderstanding: $aleph_{alpha+1}$ is not the cardinality of the power set of $aleph_alpha$; what it is is the smallest cardinal larger than $aleph_alpha$.






share|cite|improve this answer

















  • 3




    It might also be worth noting that the cardinalities obtained from $mathbb{N}$ by iterating power sets, rather than successor cardinals, are the beth numbers $beth_{alpha}$.
    – Clive Newstead
    56 mins ago










  • Thanks Clive, was literally about to comment a question about that.
    – Patrick Malone
    53 mins ago










  • @PatrickMalone I was going to reply to the comment you deleted: I sounds like you really don't know much about set theory - giving complete answers to your questions would involve writing a little essay on the topic, which is not really what MSE is supposed to be about. Lucky you, the other answer to the question is an introductory essay. You might get a lot out of amazon.com/Naive-Theory-Dover-Books-Mathematics/dp/0486814874/…
    – David C. Ullrich
    38 mins ago










  • @CliveNewstead Indeed. But then we have to try to explain why "Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?" doesn't show that $beth_{omega+1}=beth_omega$...
    – David C. Ullrich
    35 mins ago












  • The cardinality of the set representation of $omega$ is $aleph_0$, but they are not exactly equal. After all, the cardinality of the set representation of $omega +1$ is also $aleph_0$.
    – Acccumulation
    6 mins ago



















1














Ordinals are not cardinals.



Recall Hilbert's hotel. Where you have infinitely many rooms, one for each natural number, and they are all full. And there's a party, with all the guests invited. At some point, after so many drinks, people need to use the restroom.



So someone goes in, and immediately after another person comes and stands in line. They only have to wait for the person inside to come out, so they have $0$ people in front of them, and then another person comes and they only have to wait for $1$ person in front of them, and then another and another and so on. That's fine. But the person in the bathroom had passed out, unfortunately, and everyone is so polite, so they just wait quietly. And the queue gets longer.



Let's for concreteness sake, point out that only people who stay in rooms with an even room number go to the toilet. The others are just fine holding it in. Now for every given $n$, there is someone in the queue which needs to wait for at least $n$ people. The queue is infinite. But it's fine, since each person has only to wait a finite amount of time for their turn.



But what's this now? The person in room $3$ has to use the toilet as well. But they cannot cut in the line, that would be impolite. So they stand at the back. Well. There were $aleph_0$ people in the queue, that's how many, and we added just one more, so there are still $aleph_0$ people waiting in line. But now we have one person who has to wait for infinitely many people to go before them. So the queue is ordered in a brand new way. If they were lucky and someone decided to let them cut in line, then the queue would have looked the same, just from some point on people would have to wait just one more person to go first.



This is not what happened, though. So the queue looks different. Well, now we continue, all the people in room numbers which are powers of $3$ start to follow. And at some point we get to a queue which looks like two copies of the natural numbers stitched up. And then the bloke from room $5$ joins the line, and he has to two for two infinite queues before their turn. And so on and so forth.





Okay, what's the point of all that?



The point is that for finite queues the question of "how many people" and "how is the queue ordered" are the same question. So adding one person does not matter where this person was added to the queue. But when the queue was infinite, adding one person at the end or adding it to the middle would very much change the queue's order. So "how many" is no longer the same as "how long is the queue".



When iterating an operation transfinitely many times, e.g. by taking power sets or cardinal successors, we work successively. This creates a queue-like structure of cardinals. The first, the second, etc., which are ordinal numbers, they talk about order.



So once you go through the finite ones, you have to move to infinite ordinals, not to infinite cardinals. As such $omega$ is the appropriate notation, since it denotes an ordinal, rather than $aleph_0$ which denotes a cardinal.



Between $aleph_omega$ and $aleph_{omega+1}$ there are similarities: both have infinitely many [infinite] cardinals smaller than themselves. But it is not the same, exactly because we are dealing with the question "how are these ordered" rather than "how many are there".



Note that $aleph$ numbers are not defined by power sets, these are $beth$ numbers (Beth is the second letter of the Hebrew alphabet, whereas Aleph is the first one). But this is irrelevant to your actual question.






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    The reason it should really be $aleph_omega$ instead of $aleph_{aleph_0}$ is that we think of $aleph_alpha$ for ordinals $alpha$. Yes, $aleph_0=omega$, but we write $omega$ when we think of it as an ordinal instead of a cardinal.



    The rest of your question is based on a fundamental misunderstanding: $aleph_{alpha+1}$ is not the cardinality of the power set of $aleph_alpha$; what it is is the smallest cardinal larger than $aleph_alpha$.






    share|cite|improve this answer

















    • 3




      It might also be worth noting that the cardinalities obtained from $mathbb{N}$ by iterating power sets, rather than successor cardinals, are the beth numbers $beth_{alpha}$.
      – Clive Newstead
      56 mins ago










    • Thanks Clive, was literally about to comment a question about that.
      – Patrick Malone
      53 mins ago










    • @PatrickMalone I was going to reply to the comment you deleted: I sounds like you really don't know much about set theory - giving complete answers to your questions would involve writing a little essay on the topic, which is not really what MSE is supposed to be about. Lucky you, the other answer to the question is an introductory essay. You might get a lot out of amazon.com/Naive-Theory-Dover-Books-Mathematics/dp/0486814874/…
      – David C. Ullrich
      38 mins ago










    • @CliveNewstead Indeed. But then we have to try to explain why "Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?" doesn't show that $beth_{omega+1}=beth_omega$...
      – David C. Ullrich
      35 mins ago












    • The cardinality of the set representation of $omega$ is $aleph_0$, but they are not exactly equal. After all, the cardinality of the set representation of $omega +1$ is also $aleph_0$.
      – Acccumulation
      6 mins ago
















    6














    The reason it should really be $aleph_omega$ instead of $aleph_{aleph_0}$ is that we think of $aleph_alpha$ for ordinals $alpha$. Yes, $aleph_0=omega$, but we write $omega$ when we think of it as an ordinal instead of a cardinal.



    The rest of your question is based on a fundamental misunderstanding: $aleph_{alpha+1}$ is not the cardinality of the power set of $aleph_alpha$; what it is is the smallest cardinal larger than $aleph_alpha$.






    share|cite|improve this answer

















    • 3




      It might also be worth noting that the cardinalities obtained from $mathbb{N}$ by iterating power sets, rather than successor cardinals, are the beth numbers $beth_{alpha}$.
      – Clive Newstead
      56 mins ago










    • Thanks Clive, was literally about to comment a question about that.
      – Patrick Malone
      53 mins ago










    • @PatrickMalone I was going to reply to the comment you deleted: I sounds like you really don't know much about set theory - giving complete answers to your questions would involve writing a little essay on the topic, which is not really what MSE is supposed to be about. Lucky you, the other answer to the question is an introductory essay. You might get a lot out of amazon.com/Naive-Theory-Dover-Books-Mathematics/dp/0486814874/…
      – David C. Ullrich
      38 mins ago










    • @CliveNewstead Indeed. But then we have to try to explain why "Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?" doesn't show that $beth_{omega+1}=beth_omega$...
      – David C. Ullrich
      35 mins ago












    • The cardinality of the set representation of $omega$ is $aleph_0$, but they are not exactly equal. After all, the cardinality of the set representation of $omega +1$ is also $aleph_0$.
      – Acccumulation
      6 mins ago














    6












    6








    6






    The reason it should really be $aleph_omega$ instead of $aleph_{aleph_0}$ is that we think of $aleph_alpha$ for ordinals $alpha$. Yes, $aleph_0=omega$, but we write $omega$ when we think of it as an ordinal instead of a cardinal.



    The rest of your question is based on a fundamental misunderstanding: $aleph_{alpha+1}$ is not the cardinality of the power set of $aleph_alpha$; what it is is the smallest cardinal larger than $aleph_alpha$.






    share|cite|improve this answer












    The reason it should really be $aleph_omega$ instead of $aleph_{aleph_0}$ is that we think of $aleph_alpha$ for ordinals $alpha$. Yes, $aleph_0=omega$, but we write $omega$ when we think of it as an ordinal instead of a cardinal.



    The rest of your question is based on a fundamental misunderstanding: $aleph_{alpha+1}$ is not the cardinality of the power set of $aleph_alpha$; what it is is the smallest cardinal larger than $aleph_alpha$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 1 hour ago









    David C. Ullrich

    58.4k43891




    58.4k43891








    • 3




      It might also be worth noting that the cardinalities obtained from $mathbb{N}$ by iterating power sets, rather than successor cardinals, are the beth numbers $beth_{alpha}$.
      – Clive Newstead
      56 mins ago










    • Thanks Clive, was literally about to comment a question about that.
      – Patrick Malone
      53 mins ago










    • @PatrickMalone I was going to reply to the comment you deleted: I sounds like you really don't know much about set theory - giving complete answers to your questions would involve writing a little essay on the topic, which is not really what MSE is supposed to be about. Lucky you, the other answer to the question is an introductory essay. You might get a lot out of amazon.com/Naive-Theory-Dover-Books-Mathematics/dp/0486814874/…
      – David C. Ullrich
      38 mins ago










    • @CliveNewstead Indeed. But then we have to try to explain why "Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?" doesn't show that $beth_{omega+1}=beth_omega$...
      – David C. Ullrich
      35 mins ago












    • The cardinality of the set representation of $omega$ is $aleph_0$, but they are not exactly equal. After all, the cardinality of the set representation of $omega +1$ is also $aleph_0$.
      – Acccumulation
      6 mins ago














    • 3




      It might also be worth noting that the cardinalities obtained from $mathbb{N}$ by iterating power sets, rather than successor cardinals, are the beth numbers $beth_{alpha}$.
      – Clive Newstead
      56 mins ago










    • Thanks Clive, was literally about to comment a question about that.
      – Patrick Malone
      53 mins ago










    • @PatrickMalone I was going to reply to the comment you deleted: I sounds like you really don't know much about set theory - giving complete answers to your questions would involve writing a little essay on the topic, which is not really what MSE is supposed to be about. Lucky you, the other answer to the question is an introductory essay. You might get a lot out of amazon.com/Naive-Theory-Dover-Books-Mathematics/dp/0486814874/…
      – David C. Ullrich
      38 mins ago










    • @CliveNewstead Indeed. But then we have to try to explain why "Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?" doesn't show that $beth_{omega+1}=beth_omega$...
      – David C. Ullrich
      35 mins ago












    • The cardinality of the set representation of $omega$ is $aleph_0$, but they are not exactly equal. After all, the cardinality of the set representation of $omega +1$ is also $aleph_0$.
      – Acccumulation
      6 mins ago








    3




    3




    It might also be worth noting that the cardinalities obtained from $mathbb{N}$ by iterating power sets, rather than successor cardinals, are the beth numbers $beth_{alpha}$.
    – Clive Newstead
    56 mins ago




    It might also be worth noting that the cardinalities obtained from $mathbb{N}$ by iterating power sets, rather than successor cardinals, are the beth numbers $beth_{alpha}$.
    – Clive Newstead
    56 mins ago












    Thanks Clive, was literally about to comment a question about that.
    – Patrick Malone
    53 mins ago




    Thanks Clive, was literally about to comment a question about that.
    – Patrick Malone
    53 mins ago












    @PatrickMalone I was going to reply to the comment you deleted: I sounds like you really don't know much about set theory - giving complete answers to your questions would involve writing a little essay on the topic, which is not really what MSE is supposed to be about. Lucky you, the other answer to the question is an introductory essay. You might get a lot out of amazon.com/Naive-Theory-Dover-Books-Mathematics/dp/0486814874/…
    – David C. Ullrich
    38 mins ago




    @PatrickMalone I was going to reply to the comment you deleted: I sounds like you really don't know much about set theory - giving complete answers to your questions would involve writing a little essay on the topic, which is not really what MSE is supposed to be about. Lucky you, the other answer to the question is an introductory essay. You might get a lot out of amazon.com/Naive-Theory-Dover-Books-Mathematics/dp/0486814874/…
    – David C. Ullrich
    38 mins ago












    @CliveNewstead Indeed. But then we have to try to explain why "Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?" doesn't show that $beth_{omega+1}=beth_omega$...
    – David C. Ullrich
    35 mins ago






    @CliveNewstead Indeed. But then we have to try to explain why "Because if you take a powerset an infinite number of times, taking one more powerset is still just an infinite number of times, isn't it?" doesn't show that $beth_{omega+1}=beth_omega$...
    – David C. Ullrich
    35 mins ago














    The cardinality of the set representation of $omega$ is $aleph_0$, but they are not exactly equal. After all, the cardinality of the set representation of $omega +1$ is also $aleph_0$.
    – Acccumulation
    6 mins ago




    The cardinality of the set representation of $omega$ is $aleph_0$, but they are not exactly equal. After all, the cardinality of the set representation of $omega +1$ is also $aleph_0$.
    – Acccumulation
    6 mins ago











    1














    Ordinals are not cardinals.



    Recall Hilbert's hotel. Where you have infinitely many rooms, one for each natural number, and they are all full. And there's a party, with all the guests invited. At some point, after so many drinks, people need to use the restroom.



    So someone goes in, and immediately after another person comes and stands in line. They only have to wait for the person inside to come out, so they have $0$ people in front of them, and then another person comes and they only have to wait for $1$ person in front of them, and then another and another and so on. That's fine. But the person in the bathroom had passed out, unfortunately, and everyone is so polite, so they just wait quietly. And the queue gets longer.



    Let's for concreteness sake, point out that only people who stay in rooms with an even room number go to the toilet. The others are just fine holding it in. Now for every given $n$, there is someone in the queue which needs to wait for at least $n$ people. The queue is infinite. But it's fine, since each person has only to wait a finite amount of time for their turn.



    But what's this now? The person in room $3$ has to use the toilet as well. But they cannot cut in the line, that would be impolite. So they stand at the back. Well. There were $aleph_0$ people in the queue, that's how many, and we added just one more, so there are still $aleph_0$ people waiting in line. But now we have one person who has to wait for infinitely many people to go before them. So the queue is ordered in a brand new way. If they were lucky and someone decided to let them cut in line, then the queue would have looked the same, just from some point on people would have to wait just one more person to go first.



    This is not what happened, though. So the queue looks different. Well, now we continue, all the people in room numbers which are powers of $3$ start to follow. And at some point we get to a queue which looks like two copies of the natural numbers stitched up. And then the bloke from room $5$ joins the line, and he has to two for two infinite queues before their turn. And so on and so forth.





    Okay, what's the point of all that?



    The point is that for finite queues the question of "how many people" and "how is the queue ordered" are the same question. So adding one person does not matter where this person was added to the queue. But when the queue was infinite, adding one person at the end or adding it to the middle would very much change the queue's order. So "how many" is no longer the same as "how long is the queue".



    When iterating an operation transfinitely many times, e.g. by taking power sets or cardinal successors, we work successively. This creates a queue-like structure of cardinals. The first, the second, etc., which are ordinal numbers, they talk about order.



    So once you go through the finite ones, you have to move to infinite ordinals, not to infinite cardinals. As such $omega$ is the appropriate notation, since it denotes an ordinal, rather than $aleph_0$ which denotes a cardinal.



    Between $aleph_omega$ and $aleph_{omega+1}$ there are similarities: both have infinitely many [infinite] cardinals smaller than themselves. But it is not the same, exactly because we are dealing with the question "how are these ordered" rather than "how many are there".



    Note that $aleph$ numbers are not defined by power sets, these are $beth$ numbers (Beth is the second letter of the Hebrew alphabet, whereas Aleph is the first one). But this is irrelevant to your actual question.






    share|cite|improve this answer


























      1














      Ordinals are not cardinals.



      Recall Hilbert's hotel. Where you have infinitely many rooms, one for each natural number, and they are all full. And there's a party, with all the guests invited. At some point, after so many drinks, people need to use the restroom.



      So someone goes in, and immediately after another person comes and stands in line. They only have to wait for the person inside to come out, so they have $0$ people in front of them, and then another person comes and they only have to wait for $1$ person in front of them, and then another and another and so on. That's fine. But the person in the bathroom had passed out, unfortunately, and everyone is so polite, so they just wait quietly. And the queue gets longer.



      Let's for concreteness sake, point out that only people who stay in rooms with an even room number go to the toilet. The others are just fine holding it in. Now for every given $n$, there is someone in the queue which needs to wait for at least $n$ people. The queue is infinite. But it's fine, since each person has only to wait a finite amount of time for their turn.



      But what's this now? The person in room $3$ has to use the toilet as well. But they cannot cut in the line, that would be impolite. So they stand at the back. Well. There were $aleph_0$ people in the queue, that's how many, and we added just one more, so there are still $aleph_0$ people waiting in line. But now we have one person who has to wait for infinitely many people to go before them. So the queue is ordered in a brand new way. If they were lucky and someone decided to let them cut in line, then the queue would have looked the same, just from some point on people would have to wait just one more person to go first.



      This is not what happened, though. So the queue looks different. Well, now we continue, all the people in room numbers which are powers of $3$ start to follow. And at some point we get to a queue which looks like two copies of the natural numbers stitched up. And then the bloke from room $5$ joins the line, and he has to two for two infinite queues before their turn. And so on and so forth.





      Okay, what's the point of all that?



      The point is that for finite queues the question of "how many people" and "how is the queue ordered" are the same question. So adding one person does not matter where this person was added to the queue. But when the queue was infinite, adding one person at the end or adding it to the middle would very much change the queue's order. So "how many" is no longer the same as "how long is the queue".



      When iterating an operation transfinitely many times, e.g. by taking power sets or cardinal successors, we work successively. This creates a queue-like structure of cardinals. The first, the second, etc., which are ordinal numbers, they talk about order.



      So once you go through the finite ones, you have to move to infinite ordinals, not to infinite cardinals. As such $omega$ is the appropriate notation, since it denotes an ordinal, rather than $aleph_0$ which denotes a cardinal.



      Between $aleph_omega$ and $aleph_{omega+1}$ there are similarities: both have infinitely many [infinite] cardinals smaller than themselves. But it is not the same, exactly because we are dealing with the question "how are these ordered" rather than "how many are there".



      Note that $aleph$ numbers are not defined by power sets, these are $beth$ numbers (Beth is the second letter of the Hebrew alphabet, whereas Aleph is the first one). But this is irrelevant to your actual question.






      share|cite|improve this answer
























        1












        1








        1






        Ordinals are not cardinals.



        Recall Hilbert's hotel. Where you have infinitely many rooms, one for each natural number, and they are all full. And there's a party, with all the guests invited. At some point, after so many drinks, people need to use the restroom.



        So someone goes in, and immediately after another person comes and stands in line. They only have to wait for the person inside to come out, so they have $0$ people in front of them, and then another person comes and they only have to wait for $1$ person in front of them, and then another and another and so on. That's fine. But the person in the bathroom had passed out, unfortunately, and everyone is so polite, so they just wait quietly. And the queue gets longer.



        Let's for concreteness sake, point out that only people who stay in rooms with an even room number go to the toilet. The others are just fine holding it in. Now for every given $n$, there is someone in the queue which needs to wait for at least $n$ people. The queue is infinite. But it's fine, since each person has only to wait a finite amount of time for their turn.



        But what's this now? The person in room $3$ has to use the toilet as well. But they cannot cut in the line, that would be impolite. So they stand at the back. Well. There were $aleph_0$ people in the queue, that's how many, and we added just one more, so there are still $aleph_0$ people waiting in line. But now we have one person who has to wait for infinitely many people to go before them. So the queue is ordered in a brand new way. If they were lucky and someone decided to let them cut in line, then the queue would have looked the same, just from some point on people would have to wait just one more person to go first.



        This is not what happened, though. So the queue looks different. Well, now we continue, all the people in room numbers which are powers of $3$ start to follow. And at some point we get to a queue which looks like two copies of the natural numbers stitched up. And then the bloke from room $5$ joins the line, and he has to two for two infinite queues before their turn. And so on and so forth.





        Okay, what's the point of all that?



        The point is that for finite queues the question of "how many people" and "how is the queue ordered" are the same question. So adding one person does not matter where this person was added to the queue. But when the queue was infinite, adding one person at the end or adding it to the middle would very much change the queue's order. So "how many" is no longer the same as "how long is the queue".



        When iterating an operation transfinitely many times, e.g. by taking power sets or cardinal successors, we work successively. This creates a queue-like structure of cardinals. The first, the second, etc., which are ordinal numbers, they talk about order.



        So once you go through the finite ones, you have to move to infinite ordinals, not to infinite cardinals. As such $omega$ is the appropriate notation, since it denotes an ordinal, rather than $aleph_0$ which denotes a cardinal.



        Between $aleph_omega$ and $aleph_{omega+1}$ there are similarities: both have infinitely many [infinite] cardinals smaller than themselves. But it is not the same, exactly because we are dealing with the question "how are these ordered" rather than "how many are there".



        Note that $aleph$ numbers are not defined by power sets, these are $beth$ numbers (Beth is the second letter of the Hebrew alphabet, whereas Aleph is the first one). But this is irrelevant to your actual question.






        share|cite|improve this answer












        Ordinals are not cardinals.



        Recall Hilbert's hotel. Where you have infinitely many rooms, one for each natural number, and they are all full. And there's a party, with all the guests invited. At some point, after so many drinks, people need to use the restroom.



        So someone goes in, and immediately after another person comes and stands in line. They only have to wait for the person inside to come out, so they have $0$ people in front of them, and then another person comes and they only have to wait for $1$ person in front of them, and then another and another and so on. That's fine. But the person in the bathroom had passed out, unfortunately, and everyone is so polite, so they just wait quietly. And the queue gets longer.



        Let's for concreteness sake, point out that only people who stay in rooms with an even room number go to the toilet. The others are just fine holding it in. Now for every given $n$, there is someone in the queue which needs to wait for at least $n$ people. The queue is infinite. But it's fine, since each person has only to wait a finite amount of time for their turn.



        But what's this now? The person in room $3$ has to use the toilet as well. But they cannot cut in the line, that would be impolite. So they stand at the back. Well. There were $aleph_0$ people in the queue, that's how many, and we added just one more, so there are still $aleph_0$ people waiting in line. But now we have one person who has to wait for infinitely many people to go before them. So the queue is ordered in a brand new way. If they were lucky and someone decided to let them cut in line, then the queue would have looked the same, just from some point on people would have to wait just one more person to go first.



        This is not what happened, though. So the queue looks different. Well, now we continue, all the people in room numbers which are powers of $3$ start to follow. And at some point we get to a queue which looks like two copies of the natural numbers stitched up. And then the bloke from room $5$ joins the line, and he has to two for two infinite queues before their turn. And so on and so forth.





        Okay, what's the point of all that?



        The point is that for finite queues the question of "how many people" and "how is the queue ordered" are the same question. So adding one person does not matter where this person was added to the queue. But when the queue was infinite, adding one person at the end or adding it to the middle would very much change the queue's order. So "how many" is no longer the same as "how long is the queue".



        When iterating an operation transfinitely many times, e.g. by taking power sets or cardinal successors, we work successively. This creates a queue-like structure of cardinals. The first, the second, etc., which are ordinal numbers, they talk about order.



        So once you go through the finite ones, you have to move to infinite ordinals, not to infinite cardinals. As such $omega$ is the appropriate notation, since it denotes an ordinal, rather than $aleph_0$ which denotes a cardinal.



        Between $aleph_omega$ and $aleph_{omega+1}$ there are similarities: both have infinitely many [infinite] cardinals smaller than themselves. But it is not the same, exactly because we are dealing with the question "how are these ordered" rather than "how many are there".



        Note that $aleph$ numbers are not defined by power sets, these are $beth$ numbers (Beth is the second letter of the Hebrew alphabet, whereas Aleph is the first one). But this is irrelevant to your actual question.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 49 mins ago









        Asaf Karagila

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