2

I know that there is no set of all sets in ZFC. Is there a way to manipulate the axioms and define a set of all finite sets? I am asking because if you think about it, the cardinality looks a lot like a function. You give me a finite set and I can tell you a number, which is its cardinality. Input-Output. However, in order for this to be a function, it needs to have a domain. So, can this domain be defined? Or do we have to say that there cannot be a function that given a set can return its cardinality?

The curious amateur
  • 191
  • 2
    As shown in the answers to the questions linked to, the answer is no. However, this seems to be the wrong question. For instance, ${\mathbb R}$ is finite although $\mathbb R$ itself is infinite. A more sensible version would be to ask whether the collection of all hereditarily finite sets is a set. Here, $x$ is hereditarily finite if and only if $x$ is finite, and all the members of $x$ are finite, and all the members of all the members of $x$ are finite, and so on. For this version of the question, the answer is yes. – Andrés E. Caicedo May 14 '18 at 14:40
  • 1
    Let me also add to the comment by @Andrés, that the function $x\mapsto|x|$ is not a function in the sense that it is a set of ordered pair with certain properties, but it is a class function. It is a class of ordered pairs which satisfy certain properties. Moreover, it is amenable, namely if $A$ is a set, then the function $a\mapsto|a|$ for $a\in A$ is in fact a function in the traditional sense (it is a set). – Asaf Karagila May 14 '18 at 15:05

0 Answers0