# Mary Tiles confused about infinite ordinals and cardinality?

Stoney Pete
O.k. I am seriously confused... Not being to good at math but nevertheless interested in set theory, infinity, etc. I started reading Mary Tiles, The Philosophy of Set Theory (Dover edition). I particularly wanted to know more about the relation between infinite ordinals and cardinality, but what she writes about that topic is very confusing to me. Let me explain:

Starting from the following two principles Cantor was able to generate the infinite ordinals:
(1) if a is a number, then a+1 is its immediate successor;
(2) if there is a definite succession of numbers of which there is no greatest, then there is a limit or new number which is the next greatest to them all.

Using (1) you get the potential infinity of the natural numbers (0,1,2,3...). According to Tiles, Cantor called this the first number class.

Then using (2) you get the limit of the first number class which is ω, the first infinite ordinal.

Next, starting from ω and applying principles (1) and (2) over and over again, you get ω⋅2, ω⋅3... ω⋅ω... etc. According to Tiles, this is what Cantor called the second number class.

Now if I understand Tiles correctly she says that all the ordinals in the second number class have cardinality alef-0 (see below).

To get to ordinals with higher cardinality Cantor then introduced a third principle the formulation of which by Tiles remains very obscure to me but which seems to me to be an application of principle (2) to the entire succession of ordinals in the second number class. So given the succession ω, ω⋅2, ω⋅3... ω⋅ω... etc. we can (using (2)) create a limit or new number which is the next greatest, namely, ω1.

The same principles can then be appplied to ω1 and the result of this woul be the third number class whose limit is ω2 and so on...

Now comes the rub: if the cardinality of all the ordinals in the second number class is alef-0, the cardinality of the third number class (ω1, ω1+1, etc.) must be alef-1, right?

Yet this is not what Tiles says. She goes on to say some very confusing things. Here are some quotes:

"Cantor proved that the second number class cannot be put into one-one correspondence with the first...." (p.106)

This I find confusing. The first number class is simply the potential infinity of 0,1,2,3...etc. The second number class results simply from considering that potential infinity as an actual one and having its limit in ω. If we use Von Neumann's trick to define a number as the set of the numbers that precede it, then ω={0,1,2,3...etc}. Hence it follows, I would say, that ω can be put in one-one correspondence with {0,1,2,3...etc.} -- in other words, ω is denumerable. But this is not what Tiles says in the quote above.

Second problem... Tiles writes: "Thus the cardinal number of the second number class is the next after [alef-0] and is labeled [alef-1]." (p.107)

But didn't she state earlier that all the ordinals in the second number class have cardinality alef-0? This seems to follow from what she writes earlier, namely:

"All the numbers in the second number class can, however, be thought of as the numbers obtained by introducing more or less complicated order on the sequence of natural numbers... This means that although we have generated a lot of infinite ordinal numbers [...] they are all such that they are ordinal numbers of sets which can be put in one-one correspondence with the the natural numbers (denumerable sets)... So what we have is a proliferation of infinite ordinal numbers which all apply to sets having the same cardinality, [alef-0]. These first two principles [namely, (1) and (2)] on their own do not generate any ordinal number which could be the number of points in a line [i.e. alef-1]."

According to me, Tiles is very confused here. On the one hand she says that principles (1) and (2) generate the second number class and this has cardinality alef-0. But she also says that the cardinality of the second number class is alef-1!

Am I mistaken, or is Tiles contradicting herself here. Please enlighten me!

Homework Helper
The book has the word "philosophy" in it's title - so it is probably not a good source for learning about the mathematics: I'm not surprised that you find it confusing.

http://en.wikipedia.org/wiki/Aleph_number
http://en.wikipedia.org/wiki/Ordinal_number

... you may benefit from backing up a bit and looking closer at ordinals vs cardinals:
http://mathforum.org/library/drmath/view/52385.html
Starts with finite ordinals and moves quickly to the infinite ordinals - appears to deal with the specific situation where you are getting confused, but in a more straightforward language. Let me know how you get on.

Stoney Pete
Hi, thanks for your answer, the wiki-links were very useful. I think I'm getting the hang of it. I also understand what Tiles writes better now. There are, however, two questions that still bother me.

First question concerns the first number class, the natural numbers = N = 0,1,2,3.... etc. It is said that this class is not yet infinite, that ω is the first infinite ordinal, which is the initial of aleph-0. I don't get this.... How can N not be infinite if N can be put in one-one correspondence with ω since ω countable? Shouldn't N simply have aleph-0? Or is the solution here that N simply has merely potential infinity, i.e. it tends to infinity? Whereas only its limit, ω, is the first actual infinity?

Second question: How exactly is the transition made from ω to ω1 (i.e.from the second number class to the third number class)? The wiki-pages don't explain this, as far as I can see. Isn't the principle invoked here simply a version of the principle that for any succession without a greatest member you can postulate a limit, that is, a next greatest number? If you have the sequence 0,1,2,3...etc. then you can invoke this principle to postulate ω. But does this also work with the sequence ω, ω+1...ωxω, ωxωxω... etc.? So that you can postulate a limit to that sequence, which limit is then ω1 with cardinality aleph-1?

Gold Member
Hi, thanks for your answer, the wiki-links were very useful. I think I'm getting the hang of it. I also understand what Tiles writes better now. There are, however, two questions that still bother me.

First question concerns the first number class, the natural numbers = N = 0,1,2,3.... etc. It is said that this class is not yet infinite, that ω is the first infinite ordinal, which is the initial of aleph-0. I don't get this.... How can N not be infinite if N can be put in one-one correspondence with ω since ω countable? Shouldn't N simply have aleph-0? Or is the solution here that N simply has merely potential infinity, i.e. it tends to infinity? Whereas only its limit, ω, is the first actual infinity?

My guess is that what is meant is that N is finite for any given N because it has finitely many predecessors. The first infinite ordinal on the other hand has infinitely many predecessors. The principle of generation for the natural numbers is simple succession. The first infinite ordinal can be thought of as its completion.

Second question: How exactly is the transition made from ω to ω1 (i.e.from the second number class to the third number class)? The wiki-pages don't explain this, as far as I can see. Isn't the principle invoked here simply a version of the principle that for any succession without a greatest member you can postulate a limit, that is, a next greatest number? If you have the sequence 0,1,2,3...etc. then you can invoke this principle to postulate ω. But does this also work with the sequence ω, ω+1...ωxω, ωxωxω... etc.? So that you can postulate a limit to that sequence, which limit is then ω1 with cardinality aleph-1?

I am guessing again here. But once you have the first infinite ordinal then you can use both principles to generate new ordinals, simple succession and completion of simple succession. At the end of application of these two principles one has another completion, and this gives an uncountable ordinal From what you wrote it seems that for instance, simple succession and completion gets you from ω to ωxω and the again to ωxωxω and so on. The end of this application of both principles, your third principle is the first uncountable ordinal. You need to check of course that these other ones like ωxωxωxω are still countable - that is they have countably many predecessors.

What about ω^ω? Is that still countable?

Homework Helper

Stoney Pete
O.k. I think I get it now on an intuitive level....

I think the mistake I made is confusing the cardinality of the ordinals in a number class with the cardinality of the number class as a whole...

Staring with the first principle of generation:

(1) if α is an ordinal, then α+1 is the successor ordinal

you can generate the first number class (0,1,2,3... etc). Each ordinal in this class is of course finite and is its own cardinality, so e.g. card(3)=3. But to get the class as a whole you need the second principle of generation:

(2) if there is an unbounded succession of ordinals, then there is a limit ordinal which is the successor of them all

Using (2) you can complete the succession generated by (1) which gives you the first number class as a totality and this is ω. Card(ω)=aleph-0.

But ω is also the initial ordinal of the second number class. And whereas all the ordinals in that class are countably infinite, that class as a whole is uncountably infinite (aleph-0).

So how is the second number class as a whole given? Here principles (1) and (2) will not do, because they simply give you countable successors to ω. So you need a third principle to describe that class as whole. This is done by Cantor's principle of limitation which, I think, can stated roughly as follows:

(3) the second number class, whose limit is ω1, is the least greatest successor to all the ordinals with the cardinality of the first number class (aleph-0).

Or generalized (3'): the nth number class (whose limit is ωn-1) is the least greatest successor to all the ordinals with cardinality aleph-(n-1).

So given the nth number class, its initial ordinal is ωn-2 and its limit ordinal is ωn-1 and its cardinality is aleph-(n-1)...

Problem solved?

As to Lavinia's question if ω^ω is still countable... I must admit I don't what the symbol ^ means....

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Homework Helper
Part of it is the notation ... think of cardinality being the size of a set.
The set labelled "3" is, by convention, {0,1,2} so it has cardinality 3
So it makes sense to write 3={0,1,2} and card(3)=3 but it makes no sense, by this convention, to have a set labelled "0".
You can see that 3 is not a member of 3... and so on. I usually start needing a stiff drink around now.

You can probably also see how people who delve deep into this stuff tend to go a bit funny.

Note:
The symbol "^" usually means "raised to the power of". i.e. a^b would be ##a^b##

Stoney Pete
In that case: yes, ω^ω is still countably infinite

This is because it is generated by the first two principles (1) and (2) mentioned above, and anything generated by (1) and (2) is countably infinite.

Applying (1) and (2) you get in succession: ω, ω+1, ω+2...ωxω....ω^3...ω^4....ω^ω...ω^ω^ω... etc. In short you get an infinite amount of infinite ordinals, but each of these ordinals can be put in a one-to-one correspondence with ω and is therefore countably or denumerably infinite. That's quite staggering...

But it pales in comparison to the next number class which begins when the previous is gathered as one unity called ω1. Then applying (1) and (2) all over again you get: ω1+1, ω1+2... ω1^ω1... etc.

You see the ordinals and number classes never stop...

Indeed, people who work on this tend to get funny... I read that Cantor ultimately went insane from pondering these issues... But a friend of mine suggested that perhaps it was the other way around, that Cantor was born a bit loopy and hence his interest in infinity ;)