Taking the advice of @Rob Arthan, I digitised the text, corrected any errors in the digitisation before passing this text through to google translate. I then attempted to correct a couple of errors made by the machine in translation. The finished English text seems accurate to my sensibilities, however I am not in a position to give strong guarantees on the accuracy of my translation. I have been as conservative as possible in any edits I have made. Below I present the text, formatted with the usual Math-Stackechange formatting tools:
On The Notion of a Finite Set.
by Casimir Kuratowski (Warsaw).
Mr. W. Sierpiński gave in his work The axiom of M. Zermelo and its role in Set Theory and Analysis${}^1$ a new definition of the finite set. This definition is mainly distinguished by the fact that it depends neither on the notion of natural number nor on the general notion of function, which usually falls within the definitions making use of the notion of correspondence. The definition in question is as follows:
Consider class $K$ of sets each of which satisfies the following
conditions:
- Any set containing a single element is part of the class K.
- If$A$ and $B$ are two sets belonging to the class $K$, their sum-set $A + B$ also belongs to $K$.
Let us call finite any set which belongs to each of
the classes K satisfying the conditions 1 and 2.
As we know, the set of all. objects (if it exists) has paradoxical properties: contrary to a theorem known by G. Cantor, the power of this set would not be lower than that of the class of all its subsets. It is the same for the class composed of all the sets containing a single element; therefore, the classes K do not satisfy Cantor's theorem. Taking this fact into account, one could question the very existence of $K$-classes.
By modifying Mr. Sierpiński's definition so as to eliminate this drawback, I obtain the following definition:
The set $M$ is finite, when the class of all its (non-empty) subsets is
the unique class satisfying the conditions:
Its elements are (nonempty) subsets of $M$.
Any set containing a single element of $M$ belongs to this class.
If $A$ and $B$ are two sets belonging to this class, their sum-set $A + B$ also belongs to it.
We are going to demonstrate that a finite set according to this definition is also in the ordinary sense and vice versa. In other words: for a set to be finite according to the proposed definition, it is necessary and sufficient that the number of its elements can be expressed by a natural number (the notion of a natural number being assumed to be known).
Indeed, let $M$ be a set whose number of elements can be expressed by a natural number; let Z be any class satisfying conditions 1—3. We will show that every subset of $M$ belongs to $Z$. This is the case - by virtue of condition 2 - with subsets composed of a single element; at the same time, if this is the case with subsets containing $n$ elements, it is the same - according to 3 - with those which contain $n + 1$. As the number of elements of each subset of $M$ can be expressed by a natural number, it follows by induction that $Z$ contains all the subsets of $M$. Therefore, the class $Z$ being necessarily identical to that of all the subsets of $W$, it is the only class satisfying conditions 1—3. Thus, any set whose number of elements can be expressed by a natural number is a finite set in our sense.
Suppose, on the other hand, that the number of elements of a given set $M$ cannot be expressed by a natural number. Let us denote by $Z$ the class of all the subsets of $M$ whose number of elements can be expressed by a natural number. This class obviously satisfies conditions 1—3; at the same time, according to the hypothesis, $M $ does not belong to $Z$ and, therefore, $Z$ is not identical to the class of all the subsets of M; therefore, the class of all the subsets of $M$ is not the only class satisfying conditions 1 - 3 and is not finite in our sense, Q.E.D.
${}^1$ Bull. of Acad. des Sciences de Cracow, 1918, p. 106.
I have made the answer community so that if anyone confident in both English and French wants to either increase the accuracy of this translation, co-sign the accuracy of translation, or otherwise condemn it, they are free to do so.
Best Answer
Often in set theory we write $A^B$ for the collection of all functions from $B$ to $A$. Additionally by $0$ we mean $\varnothing$ and by $n$ we mean $\{0,1,\ldots,n-1\}$.
Hence in your example $f$ takes an element of $S$ to a function from $A\cup AR$ to the set $2=\{0,1\}=\{\varnothing,\{\varnothing\}\}$. By identifying characteristic functions with subsets, you can consider this as a function from $S$ to $\mathcal{P}(A\cup AR)$