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Description: Proof of Form [0 AL]

Content:

In this note we give a proof of Form [0 AL] (For every set $A$, there is an $\aleph$ which cannot be covered by fewer than $|A|$ sets each of cardinality $< |A|$.) in ZF$^0$.Form [0 AL] was suggested by K. Keremedis.

Let $A$ be a set and suppose that $K$ is an aleph which is covered by fewer than $|A|$ sets of cardinality $< |A|$. Let $\le$ be the usual ordering on $K$ and let $\bigcup_{b\in B} C_b = K$ where $|B| < |A|$ and for each $b\in B$, $|C_b| < |A|$.  Since each $C_b$ is well orderable, $|C_b| < H(A)$.  (The function $H$ is Hartogs' aleph function, $H(A)$ is the least aleph that is not $\le |A|$.) Therefore the ordering $(C_b,\le)$ has order type that of an ordinal $\alpha_b < H(A)$.  Let $f_b$ be the unique order automorphism from $\alpha_b$ onto $C_b$. Let $\gamma$ be a fixed element of $K$ and define $F : A\times H(A) \to K$ by $F(b,\beta) = f_b(\beta)$ if $b\in B$ and $\beta\in\alpha_b$ and $F(b,\beta) = \gamma$ otherwise. $F$ is onto and $|K| = |\cal E|$ where $\cal E$ is the set of equivalence classes $\subseteq A\times H(A)$ under the relation $q \sim r \Leftrightarrow F(q) = F(r)$.  Therefore $|K| = \aleph_\sigma$ where $A\times H(A)$ has a partition of cardinality $\aleph_\sigma$. Therefore, if we let $\aleph_\rho$ be the smallest well ordered cardinal greater than every cardinal in the set $\{\aleph_\sigma : A\times H(A)$ has a partition of cardinality $\aleph_\sigma\}$, then $\aleph_\rho$ cannot be covered by fewer than $|A|$ sets each of cardinality $|A|$.

Howard-Rubin number: 131

Type: Proof

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