Infinite sets without subsets of some (finite) cardinalities?

Question

While reading the Wikipedia article on infinite sets I found the following quote:

A set is infinite if and only if for every natural number, the set has a subset whose cardinality is that natural number.

If the axiom of choice holds, then a set is infinite if and only if it includes a countable infinite subset.

This raised an interesting question; without AC, is it possible to have a set where, given any natural number, you can find a subset with cardinality great than that number, but not necessary one equal to it?

For example a set where all definable subsets have even cardinality?

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Edit to clarify the question: are there sets that satisfy the intuitive meaning of infinat but not the definition as quoted above?

FWIW, id be interested in both the case where "subsets larger than n exist" requires those subsets be finite and where they don't. (Though I suspect the second case is uninteresting as most of the interesting properties would be trivially true.)

Answer 1

No, this is impossible.

If $A$ has a subset $B$ such that $B$ has size greater than $n$, then there is, by definition, an injective function $f\colon\{0,\dots,n-1\}\to B$, which is also an injective function to $A$. Taking the image of $f$ is a subset of $A$ with exactly $n$ elements.

Answer 2

No, and to be asking the question you must have misunderstood the passage you quote.

A set is infinite if and only if for every natural number, the set has a subset whose cardinality is that natural number.

There is no mention of AC here, and so this part is true even without AC.

Without AC the second part of your quote, about having a countably infinite subset, does not hold. This means you can (without AC) have an infinite set such that all its infinite subsets are uncountable, but it must still have finite subsets of every possible order.