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A set $X$ is called Dedekind-infinite if there is a injective, non-surjective map $X\to X$ and Kuratowski-infinite if $\mathcal P(X)$ is not generated by $\{\emptyset\}\cup\{\{x\}|x\in X\}$ as a sub-semilattice with respect to $\cup$.

In ZF without the axiom of infinity, any Dedekind-infinite set is also Kuratowski-infinite but the converse does not necessarily hold. However, does the existence of a Kuratowski-infinite set at least imply that a Dedekind-infinite set exists?

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Yes: if $X$ is Kuratowski-infinite, let $Y\subseteq\mathcal{P}(\mathcal{P}(X))$ be the set of equivalence classes of Kuratowski-finite subsets of $X$ with respect to cardinality. Then $Y$ is Dedekind-infinite: an injective non-surjective map $Y\to Y$ is "adding $1$" (take a representative of the equivalence class, and add one more element of $X$ which is not in it).

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Yes. Easily, a set which is not Kuratowski finite is not finite. Ant the second power set of an infinite set is Dedekind infinite.

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