Example of a set that is Dedekind-finite but not Tarski-finite?

Question

Can you give an example of a set that is Dedekind-finite, but not Tarski-finite?

Answer 1

First let me set the definitions, so there will be no fuss about the terms I use later on.

1. Finite means bijectible with a finite ordinal.
2. Dedekind-finite means "every self-injection is a bijection" (or equivalently, every self-injection is surjective).
3. Tarski-finite means that for every non-empty set of subsets has a $\subseteq$-maximal element.
4. Whenever "infinite" is being used, possibly with one of the above modifiers, the meaning is that the set does not satisfy the relevant definition of being finite (e.g. Dedekind-infinite means not Dedekind-finite).

It is provable in $\sf ZF$ that every Tarski-finite set is finite. However it is consistent with the failure of the axiom of choice that there is an infinite set which is Dedekind-finite.

But because this is a consistency result, the word "example" is misleading. Since we cannot prove, nor disprove this, from the axioms of $\sf ZF$, giving a concrete example is impossible.

In the model defined by Paul Cohen, in which the real numbers cannot be well-ordered, he does that by adding an infinite Dedekind-finite set of real numbers. Since Tarski-finiteness is the same as finiteness, this is an "example" as you are asking for. But it cannot be "given explicitly" because this is a consistency result, rather than a provability result.

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One can weaken the definition of Tarski-finite, and require that only $U$ which is linearly ordered by $\subseteq$ will have a maximal element; and under the axiom of choice this is again equivalent to being finite (and Dedekind-finite). But there are examples, similar in nature, but different from, the example I gave due to Cohen, that this weakened definition is not equivalent to being finite, or being Dedekind-finite.