\def\ac#1{#1}
\def\ap#1{}
\def\iput#1{}
\def\icopy#1{#1}
\input amstex
%\undefine\eth
\documentstyle{amsppt}
\pageheight{8.75truein}
\pagewidth{6.5truein}
\NoBlackBoxes
\loadbold
\head{NOTE 157}\endhead A proof that form 133 implies form 340.
(This proof is due to Eric Hall.)
Form 133 is the statement: Every set that cannot be well ordered
has an amorphous subset. (This is inf 2 in Rubin/Rubin [1985] where
it is shown that inf 2 is equivalent to AC in ZF, but not in ZF$^0$.)
Form 133 is true in $\Cal N1$, $\Cal N24$, $\Cal N24(n)$,
and $\Cal N26$. Form 340 is: Every Lindel\"of metric space is
separable. We shall show that 133 implies 340.
\demo{proof} Suppose $X$ with the metric $d$ is a Lindel\"of
metric space. We shall prove, assuming form 133, that $X$ is
separable. First we shall show:
\proclaim{Lemma 1} If $M$ is an amorphous subset of $X$,
then the range of $d/M\times M\ (= d[M\times M]$), is finite.
\endproclaim
\demo{Proof} For each $m\in M$, $d[{m}\times M]$ is finite
because $M$ is amorphous. Thus,
$$d[M\times M] =\bigcup_{m\in M}d[{m}\times M]$$
is a finite union of finite sets, so it is finite\enddemo
\proclaim{Lemma 2} If $X$ has an amorphous subset then $X$
is not Lindel\"of. \endproclaim
\demo{Proof} Let $M$ be an amorphous subset of $X$ and let
$S = {d(a,b): a,b\in M}$. By Lemma 1, $S$ is a finite set
and it is clearly non-empty because $M$ is infinite. Choose
a positive $\epsilon$ which is less than $\frac12$ of the
minimum number in $S$. It follows from Lemma 1 that $\epsilon$
exists.
Let $B(\epsilon,m)$, where $m\in M$, be an open neighborhood
about $m$ with radius $\epsilon$. Then
$${B(\epsilon, m): m\in M}\cup {X \setminus \overline{M}}$$
is an infinite open cover with no countable subcover, Thus,
$X$ is not Lindel\"of.\enddemo
It follows from Lemma 2 and 133 that $X$ can be well ordered.
Thus, since $X$ is a Lindel\"of metric space, it follows from
form [8 V] that $X$ is separable.\qed\enddemo
\bye