A description/proof of that collection of sets that are of non-zero cardinality is not set
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
- The reader knows a definition of cardinal number.
- The reader admits the proposition that there is no set that that contains all the sets.
Target Context
- The reader will have a description and a proof of the proposition that for any non-0 cardinal number, the collection of the sets that are of the cardinality is not any set.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any non-0 cardinal number, \(c\), the collection of sets, \(C = \{S\vert card S = c\}\) where \(card S\) is the cardinal number of \(S\) is not any set.
2: Proof
Let us suppose that \(C\) was a set. For any set, \(S'\), \(S' \in S\) for an \(S \in C\), because for any \(S \in C\), if \(S' \notin S\), there would be a set, \(S'' \in S\) because \(0 \lt card S\), and the new set, \(S''' = (S \setminus \{S''\}) \cup \{S'\}\), would be a set of the cardinality, \(c\), by the pairing axiom, the subset axiom, and the union axiom. By the union axiom, \(\cup C\), which means \(\cup_{S \in C} S\), would be a set, but that would contain all the sets, which would not be any set, by the proposition that there is no set that that contains all the sets, a contradiction.
