Weak Choice Principles and Sierpiński's Theorem: Examples and Counterexamples

Insert the MC equivalence immediately beside AC, then concatenate the proved implications through the countable pair-choice endpoint. Keep arbitrary-family pair choice on its separately proved AC branch.

The finite-level description of DC requires both DMC and countable finite choice. Thus the displayed conjunction represents exactly the proved equivalence.

If all $A_n$ are empty, the union is empty. Otherwise let $I=\{n:A_n\mathrm{\ne }\mathrm{\varnothing }\}$. Given surjections $e_n:\omega \to A_n$ for $n\in I$, enumerate $(n,k)\in I\times \omega$ by a fixed pairing enumeration and output $e_n(k)$, skipping other pairs. This lists every union member; first occurrences give an injection of the union into omega.

Without the supplied $e_n$, the set of surjections onto each nonempty $A_n$ is nonempty. For empty $A_n$ use a singleton dummy set instead. Countable choice selects these data simultaneously. The pairing operation itself costs no choice.

If the real line were such a countable union under countable choice, the union theorem would make it at most countable, contradicting its uncountability.

The first two tuple extensions are $\mathrm{\varnothing }$, $(x_0)$, $(x_0,x_1)$ with $x_0\in X_0$ and $x_1\in X_1$. In a path with one-term extensions, the $n$th tuple has length $n$; its union therefore assigns exactly one permissible value at each index in omega. This is the tuple construction in the cited implication.

For finite $R$-paths the shortest object is $(a)$. A nested path beginning at a longer object of length $m\ge 1$ has lengths $m+n$; their union still has domain omega and starts at $a$. Every adjacent pair is certified within a finite path.

Every finite subset of a partial choice graph is such a graph. Conversely, if a set of pairs is not a permissible graph, one pair has a value outside its prescribed set, or two pairs share an index with different values. A subset of size one or two witnesses the defect. The empty graph passes every finite test.

If a partial graph omits $i$, one $x\in X_i$ extends it by $(i,x)$; hence it is not maximal. A total choice graph cannot be extended permissibly, since every index already has its unique value. For empty $I$ the empty graph is already total and maximal.

Apply Specker with $X=\omega$ and the identity injection of omega. The two premises concern respectively sizes between omega and its power set, and sizes between that power set and its own power set. The conclusion is $\mathcal{P}(\omega )\approx h(\omega )$.

By the definition and characterization of the first uncountable ordinal, $h(\omega )={\omega }_1$. Substitution gives the asserted bijection, and hence well-orders the power set.

Under the stated interpretation, the equation asserts a bijection $b:\mathcal{P}({\mathrm{\aleph }}_{\alpha })\to {\mathrm{\aleph }}_{\alpha +1}$. Define $U<V$ iff $b(U)<b(V)$ in the ordinal order.

This is a linear order, and any nonempty family of subsets has a least member: take the inverse image of its image’s least ordinal. Thus the equation explicitly asserts well-orderability, refuting the claim. The equation is expressible in ZF; its being expressible does not remove this mathematical content.