Why infinite cardinalities are not “dense”?

What tells us that the structure of the cardinals is "discrete"? I'm not using the words "discrete" and "dense" with their formal meanings.

Maybe I have this confusion because I'm using concepts that I've not formalized (in my mind) but in other words why can we talk about successor of a infinite cardinal?

Who tells us that there isn't a cardinal $\kappa$ such that $\aleph_\alpha\lt\kappa\lt\aleph_\alpha^+$?

For exaple when we talk about $\mathbb Q$ we can't find the successor of a number $q^+\in\mathbb Q$ such that $\nexists r (q\lt r \lt q^+)$.

If I must define what I mean with successor I wold say this:

Let be $(A,\lt)$ a total strict order and $a\in A$ I define $succ_\lt(a)=a^+$ only if $\nexists b\in A(a\lt b\lt succ_\lt(a))$

How matematicians know that if cardinalities are linearly ordered exists a successor (like for finite sets) with the properties that I've defined? What if infinite cardinals have the structure of the rational numbers?... then we can find betwen two infinite sets always an infinite number of infinite sets with intermediate sizes? Why is not possible?

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how much do you know about ordinals? –  Ittay Weiss Apr 27 '13 at 9:38
Linked with this question How do we know an $\aleph_1$ exists at all? math.stackexchange.com/questions/46833/… –  MphLee Apr 30 '13 at 20:42

The cardinals are well-ordered. The definition you look for, by the way, is called scattered, rather than "discrete".

The reason they are well-ordered is that we use the ordinals to define the cardinals. This is a good thing, because we make a lot of use of this fact, because well-orders are easy to induct over, whereas dense orders are harder to deal with in this aspect.

We define the infinite cardinals as particular ordinals as follows:

• $\aleph_0=\omega$, the least infinite ordinal.
• $\aleph_{\alpha+1}=\omega_{\alpha+1}$ which is the least ordinal $\delta$ such that $\omega_\alpha<\delta$, and there is no bijection between the two.
• If $\aleph_\alpha$ were defined for $\alpha<\beta$ for a limit ordinal $\beta$, then $\aleph_\beta=\omega_\beta=\sup\{\omega_\alpha\mid\alpha<\beta\}$.

We have that the function $\alpha\mapsto\aleph_\alpha$ is an order-preserving bijection, so the cardinals have the same properties as the ordinals when it comes to their order.

Using the axiom of choice every set is equipotent with an ordinal, which means that every set can be assigned an $\aleph$ number. Do note that cardinality ignores any other structure you may have associated with a particular set, much like $\Bbb Q$ is countable, but there is no order-preserving bijection between $\Bbb Q$ and $\Bbb N$.

On the other hand, if we ditch the axiom of choice and replace it with its negation then not all the cardinals are ordinals and there is a lot of mess going on there. It is still not necessary that there are things which look like $\Bbb Q$, but it is consistent that there are.

With regards to the successor we have that three main definitions which are equivalent under the axiom of choice are no longer equivalent, and even $\aleph_0$ might have several distinct successors. See my answer here for more details on that last part.

But in those models of $\sf ZF$ where the axiom of choice fails, we hardly know anything about structure of the cardinals. In fact, we know very very very little on the structure of cardinals in arbitrary models of $\sf ZF$. We do know that it can get pretty darn crazy, but that's about that.

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So AC really is equivalent to "choice" a well order for the "world" of sets (because with it every set is isomorphic to an ordinal). then " It is still not necessary that there are things which look like $\Bbb Q$, but it is consistent that there are." this mean that without AC we can't say nothing about sets, because how are cardinalities ordered is "undecidable" (let me use this word without a study of model theory)? –  MphLee Apr 30 '13 at 16:54
@MphLee: No, that's global choice. The axiom of choice is a "local" version which states that every set has its own well-ordering; you don't necessarily have the ability to glue them all together. And in a nutshell yes, without the axiom of choice we just know that the cardinals are not linearly ordered and that they have certain pathologies. It is still not known, for example, if the failure of AC implies there is a decreasing chain of cardinals or not. So we really don't know much, but we can engineer models of set theory where the structure is quite chaotic. –  Asaf Karagila Apr 30 '13 at 18:23

Assuming Choice it follows that there is an order-preserving bijection between the ordinal numbers and the (infinite) cardinals. (This is where the notations $\aleph_0 , \aleph_1 , \ldots , \aleph_\alpha , \ldots$ comes from.) As every set of ordinal numbers has a least element, the same is true about sets of cardinal numbers.

In particular, given any cardinal $\kappa$ as $\kappa < 2^\kappa$ we can consider the set $\{ \lambda : \kappa < \lambda \leq 2^\kappa \}$, and take the least element of this set to get the smallest cardinal number greater than $\kappa$.

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If I may add, under AC assumption, there exists a successor over $\mathbb Q$, even over $\mathbb R$. These are simply not compatible with the usual linear order that we define on these sets.

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Note that $\Bbb Q$ is countable without using choice, so a successor exists anyway. But more subtly without the axiom of choice the definition of successor can be given in three non-equivalent ways. See my answer here. –  Asaf Karagila Apr 27 '13 at 9:57
@AsafKaragila Very Intresting but really I'dont get how can you find the "exact" value of a $\lt$-successor on $\mathbb Q$ ($\lt$ is the usual order on the rationals). For me is ..impossible to find a fraction like that. (ex. the $succ_\lt(0)=0+(1/n)$ with $n$ "infinite"... :S) –  MphLee Apr 27 '13 at 17:15
@MphLee: Dense orders don't have a successor function. The cardinals are not ordered densely, and when they do then the successor function is not defined everywhere. –  Asaf Karagila Apr 27 '13 at 17:18