Description:
Form 9 (Dedekind finite = finite) is equivalent to
[9 C] (The image of a Dedekind finite set is
Dedekind finite.)
Content:
Form 9 (Dedekind finite = finite) is equivalent to
[9 C] (The image of a Dedekind finite set is
Dedekind finite.)
Clearly Form 9 implies [9 C]. By
Note 8 (above) to prove [9 C] implies 9, it suffices to prove
[9 C] implies [9 A]. Assume
[9 C] and that \(x\) is an infinite, Dedekind finite set and that \(\cal P(x)\) is
Dedekind infinite. Then there is a one to one function \(f: \omega\rightarrow\cal P(x)\).
We may assume that \((\forall i,j\in\omega)(i\neq j\)implies \(f(i)\cap f(j) = \emptyset)\).
If \(g: \omega\rightarrow\cal P(x)\) is one to one, then we define \(f: \omega\rightarrow\cal P(x)\) by induction so that for each
\(i\in\omega\), \(f(i)\cap f(j) = \emptyset\) for all \(j < i\), and such that the set algebra generated by
\(\{g(k)\cap (x-\bigcup^{}_{j\le i}f(j)): k > j\}\) is infinite. \(f(0)\) is defined as follows: Either the set algebra generated by
\(\{g(k)\cap g(0): k > 0\}\) is infinite or the set algebra generated by \(\{g(k)\cap (x - g(0)): k > 0\}\) is infinite. If the latter
is true then \(f(0) = g(0)\). If the latter is false then \(f(0) = x- g(0)\). Suppose that \(f(i)\) has been defined for every \(i < n\)
satisfying the required conditions. Then for some \(k\) (and we take the least such) \( g(k)\cap (x-\bigcup^{}_{j < n} f(j))\neq\emptyset\)
(call this set \(A\)) and \((x - \bigcup^{}_{j < n} f(j)) - g(k)\neq\emptyset\) (call this set \(B\)). Since \(A\cup B=(x - \bigcup^{}_{j < n} f(j))\),
one of the two set algebras, either that generated by \(\{g(k)\cap A: k > n \}\) or that generated by \(\{g(k)\cap B: k > n\}\), is infinite.
If the latter is the case, let \(f(k) = A\). If the latter fails, let \(f(k) = B\). This completes the proof of the lemma.
Now define \(h: X \rightarrow{\cal P}(x)\) by
\[h(t) = \left\{ \begin{array}{lp{0.1}l}
f(i), & & \hbox{if } t \in f(i) \\
\emptyset, & & \hbox{otherwise}
\end{array}
\right.
\]
By [9 C] the image of \(X\) under \(h\) is Dedekind finite, a contradiction.
Howard-Rubin number:
11
Type:
proof of equivalencies
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