How should we parse this proof that every infinite set has a subset which is eqivalent to the set of natural numbers?

Question

This is again from the chapter Construction of the System of Real Numbers in The Fundamentals of Mathematics, Volume 1. It may be the case that the original wording in the German language would be easier to follow. As it stands, I'm not exactly sure how to break down the following (half) paragraph.

In general, two sets are said to be equivalent if either of them can be mapped one-to-one onto the other. Thus the finite sets are defined as those sets that are equivalent to the segments $A_{n}.$ For convenience, the empty set $\emptyset,$ which contains no elements at all, is also said to be a finite set. In an infinite set $M,$ namely a set which is not finite, it is easy to determine a set which is equivalent to the set $\mathbb{N}$ of all natural numbers: for if $f$ is a mapping which to each non-empty subset $X$ of the set $M$ assigns an element of $f\left[X\right]$ of the subset $X,$ the sets $M_{1},M_{2},\dots$ can be defined recursively by $M_{1}=\left\{ f\left[M\right]\right\} ,$$M_{n^{\prime}}=M_{n}\cup\left\{ f\left[M-M_{n}\right]\right\} ,$ and then the union of the $M_{n}$ provides us with the desired subset $\mathbb{N^{*}}.$

The part I'm not sure about is:

$\dots$ if $f$ is a mapping which to each non-empty subset $X$ of the set $M$ assigns an element of $f\left[X\right]$ of the subset $X,$ $\dots$

In particular, am I supposed to understand $f\left[X\right]$ to be a set? That's what I would typically make of the phrase "an element of $f\left[X\right]$." But that would mean, for example, that $M_{1}=\left\{ f\left[M\right]\right\}$ is a nested set, and therefore $\left[M-M_{1}\right]=M$ since $M$ has no elements which are sets.

Could the "of" in "an element of $f\left[X\right]$" be dropped or replaced by "designated by", or something similar?

It is also unclear what it means to assign $f\left[X\right]$ to the set $X$.

How might we reword the entire passage:

if $f$ is a mapping which to each non-empty subset $X$ of the set $M$ assigns an element of $f\left[X\right]$ of the subset $X,$ the sets $M_{1},M_{2},\dots$ can be defined recursively by $M_{1}=\left\{ f\left[M\right]\right\} ,$ $M_{n^{\prime}}=M_{n}\cup\left\{ f\left[M-M_{n}\right]\right\} ,$ and then the union of the $M_{n}$ provides us with the desired subset $\mathbb{N^{*}}.$,

to be clearer? I'm trying, but it isn't as easy as it seems.

Answer 1

Is this your own translation? To get it to make sense, all you need to do is replace

an element of $f[X]$ of the subset $X$

with

an element $f[X]$ of the subset $X$

Answer 2

$f[X]$ is an element of $M$ (when $X$ is a nonempty subset of $M$). That is, $$f:\mathcal{P}_{\not=\emptyset}(M)\rightarrow M.$$ This means that $M-M_1$ is not just $M$, since $M=\{f[M]\}$ is a set containing a single element of $M$.

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I think it's helpful to first consider a specific example: if we take $M=\{1,2,3,...\}$ and $f[X]=\min(X)$ (note that this fits the rule: $f$ is indeed defined on every nonempty subset of $M$, and sends a nonempty subset of $M$ to an element of that subset), then:

• $M_1=\{f[M]\}=\{\min(\{1,2,3,...\})\}=\{1\}$.

• $M_2=M_1\cup \{f[M\setminus M_1]\}=\{1\}\cup\min(\{2,3,4,...\})=\{1,2\}.$

• And in general, $M_k=\{1,2,...,k\}$.