# Is $\aleph_0$ the minimum infinite cardinal number in $ZF$?

$\aleph_0$ is the least infinite cardinal number in ZFC. However, without AC, not every set is well-ordered.

So is it consistent that a set is infinite but not $\ge \aleph_0$? In other words, is it possible that exist an infinite set $A$ with hartog number $h(A)=\aleph_0$?

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It is certainly consistent that a set $S$ exists that is not finite, but such that no injetion $f:\mathbb{N}\to S$ exists. –  Michael Greinecker Nov 29 '12 at 10:34

## 1 Answer

Yes. It is consistent to have such sets. These sets are called Dedekind-finite sets.

Do note that $\aleph_0$ is the only minimal infinite cardinal, even if it not the minimum.

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Yes, exactly $A$ is Dedekind-infinite iff $\aleph_0 \le |A|$. –  Popopo Nov 29 '12 at 11:18