Exploring The Relation Between Cardinal Sum And Cardinal Power In Set Theory

In the fascinating realm of set theory, particularly when we venture into the landscape of choiceless set theory, the interplay between cardinal arithmetic operations reveals intriguing relationships. This article delves into the connection between two seemingly distinct statements involving cardinal numbers: the assertion that for an infinite cardinal $\mathfrak{a}$, $\mathfrak{a} + \mathfrak{a} = \mathfrak{a}$, and the claim that $2^{\mathfrak{a}} = \mathfrak{a}^{\mathfrak{a}}$. Understanding this relationship requires us to navigate the intricacies of cardinal addition, cardinal exponentiation, and the subtle role played by the Axiom of Choice.

Before diving into the specifics, let's establish the fundamental concepts of cardinal arithmetic. A cardinal number is a measure of the 'size' of a set. For finite sets, the cardinal number is simply the number of elements. However, when dealing with infinite sets, we encounter the transfinite cardinals, denoted by symbols like $\aleph_0$ (the cardinality of the set of natural numbers) and $\mathfrak{c}$ (the cardinality of the continuum, the set of real numbers). Cardinal addition and exponentiation extend the familiar arithmetic operations to these transfinite cardinals.

Cardinal addition is defined as follows: If $\mathfrak{a}$ and $\mathfrak{b}$ are cardinal numbers, then $\mathfrak{a} + \mathfrak{b}$ is the cardinality of the union of two disjoint sets $A$ and $B$ with cardinalities $\mathfrak{a}$ and $\mathfrak{b}$, respectively. In simpler terms, you take two sets of the given sizes, ensure they have no elements in common, combine them, and then find the size of the resulting set. A crucial result in cardinal arithmetic is that for any infinite cardinal $\mathfrak{a}$, $\mathfrak{a} + \mathfrak{a} = \mathfrak{a}$ if we assume the Axiom of Choice. This might seem counterintuitive at first, but it means that combining two sets of the same infinite size doesn't actually result in a larger set.

Cardinal exponentiation is defined similarly: $\mathfrak{a}^{\mathfrak{b}}$ is the cardinality of the set of all functions from a set of cardinality $\mathfrak{b}$ to a set of cardinality $\mathfrak{a}$. This can be thought of as the number of ways to map a set of size $\mathfrak{b}$ into a set of size $\mathfrak{a}$. A particularly important case is $2^{\mathfrak{a}}$, which represents the cardinality of the power set of a set with cardinality $\mathfrak{a}$ (the power set is the set of all subsets). Cantor's theorem establishes a fundamental result: for any cardinal $\mathfrak{a}$, $2^{\mathfrak{a}} > \mathfrak{a}$. This tells us that the power set of any set is always strictly larger than the original set itself.

Cardinal arithmetic behaves differently than standard arithmetic, particularly with infinite cardinals. The equation $\mathfrak{a} + \mathfrak{a} = \mathfrak{a}$ holding for infinite cardinals is a hallmark of this difference, and its relationship with other cardinal properties, like exponentiation, is a key area of study in set theory.

Let's formally define the two statements we're interested in:

• $\varphi$: If $\mathfrak{a}$ is an infinite cardinal, then $\mathfrak{a} + \mathfrak{a} = \mathfrak{a}$.
• $\psi$: If $\mathfrak{a}$ is an infinite cardinal, then $2^{\mathfrak{a}} = \mathfrak{a}^{\mathfrak{a}}$.

The statement $\varphi$ essentially says that adding an infinite cardinal to itself doesn't change its size. This is a cornerstone of cardinal arithmetic when the Axiom of Choice is assumed. However, without the Axiom of Choice, this statement is not necessarily true.

The statement $\psi$ relates the power set cardinality ($2^{\mathfrak{a}}$) to the cardinality of all functions from a set of size $\mathfrak{a}$ to itself ($\mathfrak{a}^{\mathfrak{a}}$). Intuitively, $\mathfrak{a}^{\mathfrak{a}}$ represents the number of functions from a set of cardinality $\mathfrak{a}$ to itself. This value can be vastly larger than $\mathfrak{a}$ depending on the cardinality of $\mathfrak{a}$.

The question we aim to address is: What is the relationship between these two statements? Does one imply the other? Are they equivalent? The answer lies in the nuances of cardinal arithmetic and the subtle role played by the Axiom of Choice.

The Axiom of Choice (AC) is a fundamental principle in set theory that states that for any collection of non-empty sets, it is possible to choose one element from each set. While seemingly intuitive, AC has profound implications and consequences, some of which are quite surprising. In particular, AC dramatically simplifies cardinal arithmetic. With AC, the sum of two infinite cardinals is simply the larger of the two, and as we mentioned before, $\mathfrak{a} + \mathfrak{a} = \mathfrak{a}$ for any infinite cardinal $\mathfrak{a}$.

However, without the Axiom of Choice, the situation becomes much more complex. It is possible to have infinite cardinals $\mathfrak{a}$ for which $\mathfrak{a} + \mathfrak{a} > \mathfrak{a}$. This is one of the reasons why choiceless set theory is so fascinating – it challenges our intuitions about the sizes of infinite sets.

When we consider the relationship between $\varphi$ and $\psi$, the Axiom of Choice (or its absence) becomes a crucial factor. Under AC, both statements are generally considered true within the standard framework of ZFC (Zermelo-Fraenkel set theory with the Axiom of Choice). Without AC, the landscape changes significantly, and the relationship between the statements requires more careful analysis.

Let's examine the relationship between $\varphi$ and $\psi$ more closely. We want to determine if one statement implies the other, and whether they might be equivalent.

$\varphi$ implies $\psi$: To show that if $\mathfrak{a} + \mathfrak{a} = \mathfrak{a}$ then $2^{\mathfrak{a}} = \mathfrak{a}^{\mathfrak{a}}$, we can proceed as follows. Recall that $\mathfrak{a}^{\mathfrak{a}}$ represents the cardinality of the set of all functions from a set of size $\mathfrak{a}$ to itself. We can represent each such function as a set of ordered pairs. Since $\mathfrak{a}$ is infinite, $\mathfrak{a} \times \mathfrak{a}$ has the same cardinality as $\mathfrak{a}$ if $\mathfrak{a} + \mathfrak{a} = \mathfrak{a}$. Now, consider a function from $\mathfrak{a}$ to {0, 1}, where {0,1} represents the cardinal number 2. The number of such functions is $2^{\mathfrak{a}}$. Any function from $\mathfrak{a}$ to $\mathfrak{a}$ can be viewed as a subset of $\mathfrak{a} \times \mathfrak{a}$, so the cardinality of the set of functions from $\mathfrak{a}$ to $\mathfrak{a}$, which is $\mathfrak{a}^{\mathfrak{a}}$, is less than or equal to the cardinality of the power set of $\mathfrak{a} \times \mathfrak{a}$, which is $2^{\mathfrak{a} \times \mathfrak{a}}$. Since $\mathfrak{a} \times \mathfrak{a}$ has the same cardinality as $\mathfrak{a}$, we have $2^{\mathfrak{a} \times \mathfrak{a}} = 2^{\mathfrak{a}}$. Therefore, $\mathfrak{a}^{\mathfrak{a}} \leq 2^{\mathfrak{a}}$.

Now, consider a function f from $\mathfrak{a}$ to {0, 1}. There are $2^\mathfrak{a}$ such functions. We can also view this as a function from $\mathfrak{a}$ to $\mathfrak{a}$, since {0, 1} is a subset of $\mathfrak{a}$ when $\mathfrak{a}$ is infinite. Thus, the number of functions from $\mathfrak{a}$ to {0, 1} is less than or equal to the number of functions from $\mathfrak{a}$ to $\mathfrak{a}$, so $2^{\mathfrak{a}} \leq \mathfrak{a}^{\mathfrak{a}}$.

Combining these inequalities, we have $\mathfrak{a}^{\mathfrak{a}} \leq 2^{\mathfrak{a}}$ and $2^{\mathfrak{a}} \leq \mathfrak{a}^{\mathfrak{a}}$, which implies $2^{\mathfrak{a}} = \mathfrak{a}^{\mathfrak{a}}$. This demonstrates that $\varphi$ implies $\psi$.

$\psi$ implies $\varphi$: The converse, whether $2^{\mathfrak{a}} = \mathfrak{a}^{\mathfrak{a}}$ implies $\mathfrak{a} + \mathfrak{a} = \mathfrak{a}$, is more challenging to prove without the Axiom of Choice. It is not generally true in choiceless set theory. There are models of set theory without AC where $2^{\mathfrak{a}} = \mathfrak{a}^{\mathfrak{a}}$ holds for some infinite cardinal $\mathfrak{a}$, but $\mathfrak{a} + \mathfrak{a} > \mathfrak{a}$. This means that statement $\psi$ does not necessarily imply statement $\varphi$ in the absence of the Axiom of Choice.

The existence of models where $\psi$ holds but $\varphi$ does not highlights the subtle nature of cardinal arithmetic in choiceless set theory. These models are often constructed using advanced techniques in set theory, and their existence demonstrates that the Axiom of Choice plays a crucial role in establishing the familiar properties of cardinal arithmetic.

In these models, it's possible to have infinite cardinals that behave in ways that are impossible under AC. For example, it's possible to have a cardinal $\mathfrak{a}$ such that $\mathfrak{a} < 2^{\mathfrak{a}} < 2^{2^{\mathfrak{a}}}$, yet $\mathfrak{a} + \mathfrak{a} > \mathfrak{a}$. These counterintuitive results underscore the importance of carefully considering the foundational axioms when working with infinite sets.

In summary, we've explored the relationship between the statements $\varphi$ (if $\mathfrak{a}$ is an infinite cardinal then $\mathfrak{a} + \mathfrak{a} = \mathfrak{a}$) and $\psi$ (if $\mathfrak{a}$ is an infinite cardinal then $2^{\mathfrak{a}} = \mathfrak{a}^{\mathfrak{a}}$). We've shown that $\varphi$ implies $\psi$, but the converse is not necessarily true in the absence of the Axiom of Choice. This exploration highlights the critical role of the Axiom of Choice in cardinal arithmetic and demonstrates the fascinating complexities that arise when we venture into choiceless set theory.

The world of infinite sets is vast and often counterintuitive. Understanding the interplay between basic operations like addition and exponentiation requires a careful consideration of the underlying axioms and the logical framework within which we operate. The relationship between $\varphi$ and $\psi$ serves as a compelling example of the subtle but profound impact of the Axiom of Choice on our understanding of infinity.