In $\mathbf{ZF}$, it is possible for a set $A$ to be infinite but not to admit a countable set. In other words, for any $\alpha\in\omega$, there is an injection from $\alpha$ into $A$, but there is no injection from $\omega$ into $A$. If we replace $\omega$ by a successor cardinal $\kappa^+$ in the above statement, any $A$ of cardinality $\kappa$ works, so there exists an example in $\mathbf{ZFC}$. However, for limit cardinals, the question seems unclear to me.


Therefore: If $\kappa>\omega$ is a limit cardinal and $A$ is a set such that for any $\alpha\in\kappa$, there is an injection from $\alpha$ into $A$, does it follow in $\mathbf{ZF}$ that there is an injection from $\kappa$ into $A$? Additionally, does the existence of such an $A$ imply the existence of a Dedekind-finite set?



Dedekind-"finiteness" for arbitrary limit cardinals
Start with your favourite model of $\sf ZFC$, your favourite regular cardinal $\mu$, and your favourite limit cardinal $\lambda>2^\mu$.


Now consider the ${<}\mu$-support product $\prod\_{\alpha<\lambda}\operatorname{Add}(\mu,\alpha)$. With automorphism groups that act on each individual component in the product, and with a filter of subgroups generated by fixing a bounded1 number of component pointwise, we get a symmetric system. It is easy to see that in the symmetric extension $2^\mu$ has an injection from any $\alpha<\mu$, since that is given by the $\alpha$th component.


But if there is an injection from $\lambda$ into $2^\mu$, then this injection codes a set of ordinals, and by a standard homogeneity argument we can show that any set of ordinals added to the model must have been added by a bounded part of the product, and so that is impossible.


Now, since this forcing is $\mu$-closed, has $\mu^+$-cc and the filter of groups has $\operatorname{cf}(\lambda)$-completeness we instantly get $\sf DC\_{<\operatorname{cf}(\lambda)}$ to hold in the extension as well.


Picking $\mu=\omega$ and $\lambda=\beth\_{\omega\_1}$, for example, provides us with $\sf DC$, and with $\aleph(\Bbb R)=\beth\_{\omega\_1}$.




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1. We can actually replace bounded by any other support, as long as it does not contain any cofinal subsets. This will affect how much $\sf DC$ holds in your model, though.


