# Question about Halmos' Naive Set Theory

Halmos proves shortly before the cited paragraph that finite subsets are not equivalent to themselves. He then says the following:

The number of elements in a finite set E is, by definition, the unique natural number equivalent to E; we shall denote it by #(E). It is clear that if the correspondence between E and #(E) is restricted to the finite subsets of some set X, the result is a function from a subset of the power set $\mathcal{P}(x)$ to $\omega$.

This is all clear (a specification on dom[$E \rightarrow \#(E)$] of $s : s \subset X$). He then continues:

This function is pleasantly related to the familiar set-theoretic relations and operations. Thus, for example, if $E$ and $F$ are finite sets such that $E \subset F$, then $\#(E) \leq \#(F)$. (The reason is that since $E \cong \#(E)$ and $F \cong \#(F)$, it follows that $\#(E)$ is equivalent to a subset of $\#(F)$.)

I do not follow his reasoning. The fact itself is clear, but I do not see the implication which he is suggesting. Is he indicating that $E$, as a subset of $F$, will be mapped by this function to some natural number which is a subset of $\#(F)$?

Edit: The passage continues:

Another example is the assertion that if $E$ and $F$ are finite sets, then $E \bigcup F$ is finite, and, moreover, if $E$ and $F$ are disjoint, then $\#(E \bigcup F) = \#(E) + \#(F).$ The crucial step in the proof is the fact that if $m$ and $n$ are natural numbers, then the complement of $m$ in the sum $m+n$ is equivalent to $n$; the proof of this auxiliary fact is achieved by induction on $n$. Similar techniques prove that if $E$ and $F$ are finite sets, then so also are $E \times F$ and $E^F$, and, moreover, $\#(E \times F) = \#(E) * \#(F)$ and $\#(E^F) = \#(E)^{\#(F)}$.

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By definition $\#(E)\cong E\subseteq F\cong \#(F)$. The equivalence $F\cong\#(F)$ means that there is a bijection $h:F\to\#(F)$; clearly $h[E]\subseteq\#(F)$. The equivalence $E\cong\#(E)$ means that there is a bijection $g:\#(E)\to E$. The composition $h\circ g:\#(E)\to\#(F)$ maps $\#(E)$ bijectively to $$h\big[g[\#(E)]\big]=h[E]\subseteq\#(F)\;,$$ thereby establishing that $\#(E)\cong h[E]$, which is a subset of $\#(F)$.
He’s not saying that $\#(E)$ is a subset of $\#(F)$, just that it’s equivalent to one.
@user1296727: In this case he appears to be saying that the subset relation between $E$ and $F$ is (almost) mirrored by the $\#$ function: you don’t quite get $\#(E)\subseteq\#(F)$, but you do get $\#(E)$ equivalent to a subset of $\#(F)$. Similarly, you’ll have $\#(A)$ and $\#(B)$ equivalent to subsets of $\#(A\cup B)$. I’d have to read the whole passage to see whether he actually does much with this, or whether it’s more of an aside, and I don’t have the book. – Brian M. Scott Jun 4 '12 at 4:05