Nonstandard cardinalities and $\mathbb{N}$

Question

Let us work in the standard ZFC universe $S$. Within it there also exist nonstandard models of ZFC. In our standard universe we have (standard) $\mathbb{N}$ with countable cardinality. Within the nonstandard model $L$ of ZFC we can have a nonstandard model $K$ of Peano Arithmetic that has some uncountable cardinality by the theorem of Lowenheim and Skolem.

But because the nonstandard ZFC model is within our standard model of ZFC, we can find an injective function $f$ that maps the standard set $\mathbb{N}$ to $K$. What cardinality does the image $I$ of $f$ have with respect to the nonstandard ZFC model $L$?

From what I have been able to find [Theorem 2 and following text, page 663, here], $I$ is not actually regarded as a valid set in $K$. Why is this -- is $\in^L$ different from $\in^S$? Is $\in^L$ some complex operator or is it merely $\in^S$ with added stipulations?

Or, does $I$ have some nonstandard finite cardinality in $K$, thus appearing finite with respect to $L$? What does it even mean to have a nonstandard cardinality? For example, if $\mathbb{N}$ has nonstandard cardinality $H$, then what about $H-1$? Something must have cardinality $H-1$, but then this would be a set that is not, from the perspective of the standard universe, finite; but it has cardinality less than $\mathbb{N}$.

Answer 1

The injection $f$ is an element of $V$, not $L$ (incidentally, you really shouldn't use "$L$" to denote a nonstandard model of ZFC). There is no reason to believe that its image is in $L$ at all. Its image will be a subset of $L$, but not all subsets of $L$ correspond to elements of $L$. (Brute force way to see this: in $V$, $L$ has $2^{\vert L\vert}$-many subsets, but clearly only $\vert L\vert$-many of them can correspond to elements of $L$.)

Basically: whenever you have some object defined by referencing $V$, there is no reason to expect that it or anything defined in terms of it should live inside your other model.

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To be absolutely precise: there is no reason to believe that there is some $x\in L$ such that $$\{y\in L: L\models y\in x\}=im(f).$$ (Incidentally, if you prefer we can write "$\{y\in L: L\models y\in x\}$" as "$\{y\in^V L: y\in^L x\}$.")