About C. R. Acad. Sci. Paris 236 (1953), 265–268.
I obtained it by asking "Gallica" and then the full title.
I have translated (see below) a part of the document.
Beginning of the document's translation:
SET THEORY : About stationary sets of ordinal numbers and distinguished sequences of regressive functions. Note by Mr. Gérard Bloch, presented by Mr. Arnaud Denjoy.
On the set of ordinal numbers of class I and II, a family of subsets, said “stationary” is defined and studied ; the results are then applied to distinguished sequences associated with ordinal numbers of second kind.
O denotes the set of cardinals of classes I and II with order topology. $\Omega$ is the first number of class III. Please note that the intersection of a denumerable family of non denumerable closed sets is a non denumerable closed set.
A function $f(x)$, defined on a subset $A$ of $O$ taking its values in $O$, will be said regressive if $f(x)\leq x$ for all x in A with equality excluded but for $x=1$, if $1 \in A$. A set $A$ of points belonging to $\Omega$ is stationary if it is non denumerable and if for any regressive function $f(x)$ defined on $A$, one has $\lim \cdots \Omega$; otherwise said if there exists at least a point a such that $f^{-1}(a)$ is non denumerable. In all other cases, A is non stationary.
Theorem 1. _ A necessary and sufficient condition of stationarity for a given set is that any closed subset of its complement set is at most denumerable.
In particular, any set containing a closed non denumerable subset (in particular the whole set) is stationary. Moreover, any set which is the union of a denumerable family of non-stationary sets is itself non-stationary; and the intersection of a non denumerable closed set and a stationary set is a stationary set.
Let $O’$ denote the set of numbers of the second type, a distinguished sequence over a subset $A$ of $ O'$ will be by definition a sequence of regressive functions $f_n(x) \ n \in \mathbb{N}$ defined on $A$, such that, for all $x \in A$
$f_1(x)\leq f_2(x)\leq\cdots$ and $\lim_{n} f_n(x)=x$.
Theorem II:- A distinguished sequence being given over a stationary subset $A$ of $O’$, there exists an integer $n_0$ such that for any $n>n_0$, and for any partition of $A$ into two subsets $B_n$ and $C_n$ such that $f_n(x)$ is upper bounded on $B_n$, the set $C_n$ is stationary.
The proof is omitted by lack of place.
By suppressing then, if needed, a finite number of terms in the distinguished sequence we may assume that $n_0=i$. Consider then function $f(x)$ and the set of points of $O$ whose reciprocal image by $f_i(x)$ is non denumerable. $H$ is itself non denumerable. Indeed if $K=f_i^{-1}(H)$ and $L=A-K$, set $L$ is not stationary; were $H$ be denumerable, $f_i(x)$ would be bounded from above on $K$ and theorem II would be invalidated. Therefore, one can number the elements of $h_i$ of $H$ by increasing order of magnitude, index $i$ describing the whole space $O$.
[...]
Corollary : any stationary set is the union of two stationary disjoint sets ; any non denumerable closed set (in particular the whole space) is the union of two disjoint sets none of which contains a non denumerable closed set.
By the way this corollary is a particular case of the following theorem which can be proved directly : if, on a set E whose cardinality is at most the cardinality of continuum, a family $F$ of sets, closed for operation “denumerable intersection”, is given, there exists a partition of $E$ into two disjoints sets, none of which contains a set of $F$.
[...]
Theorem III : On any closed non denumerable set $F$ (especially on the set $O’$ of numbers of the second category), there exists a distinguished sequence $f_1(x) \cdots f_n(x)…$ with the following properties :
1) The set F is the union of $\aleph_1$ stationary sets pairwise without common points, on each of which $f_i(x)$ takes a constant value, all of these values being different.
2) Each set of the $p$th order is the union of $\aleph_1$ stationary set pairwise without common points, on which each $f_{p+1}(x)$ takes a constant values, all of them different. These sets will be called “sets of $(p+1)$st order”.
3) The intersection of the sequence of sets $E_1,E_2\cdots$ such that $E_p$ is a set of $p$th order” with $E_1 \supset E_2\cdots \supset \cdots$ is void, or reduced to a point $x$. In the latter case, $x$ is the limit of increasing sequence $x_1,x_2, \cdots$, where $x_p$ denotes the value of $f_p(x)$ on $E_p$.
In a reverse way, any point $x \in F$ is the intersection of such a sequence of sets $E_p$, and it it possible to prove that the set of $x$ elements which are not equal to their rank in at least one of the sets $E_p$ containing it is a non stationary set.
Note : Mr. Neumer Walter has presented, in the case of the set of ordinal numbers strictly less than an initial given number of classes, a part of the previous results dealing with stationary sets, but for results dealing with “distinguished sequences"(Cf. Math. Zeitung, 54, p. 254-261). This article was unknown to me when I wrote this Note.
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Best Answer
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:
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:
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.