Let $V$ be the class of all sets, $\text{Ord}$ be the class of all ordinals, $\text{InOrd}$ be the class of all initial ordinals, and $f:V \to \text{Ord}$ be the function that sends each set to its corresponding Hartogs number.
As a result, $f(a)=a+1$ for all $a \in \Bbb N$ and thus $f \restriction \Bbb N$ is injective.
I find it difficult to imagine the Hartogs numbers of infinite sets.
My question: Are $f,f\restriction \text{Ord},f\restriction \text{InOrd}$ injective?
Thank you in advance for your help!
The answers to $1$ and $2$ are no: for example, it's easy to show that $f(\omega)=f(\omega+1)=\omega_1$ (where $\omega_1$ is the first uncountable ordinal). More generally, what can you say about $f(x)$ and $f(y)$ if there is a bijection between $x$ and $y$ (that is, if two sets have the same cardinality, how are their Hartogs numbers related)?
The connection with cardinality, in turn, suggests that the situation with respect to cardinals (= initial ordinals) might be different. And indeed it is:
Suppose $\kappa<\lambda$ are infinite cardinals and $f(\kappa)=f(\lambda)$.
We clearly have $\lambda<f(\lambda)$ (since there's a bijection between $\lambda$ and $\lambda+1$), so we get $\lambda<f(\kappa)$.
But then there must be a surjection from $\kappa$ to $\lambda$. So $f$ is injective on the initial ordinals.
Note that as long as we have the axiom of choice, the Hartogs number is only about cardinality: $f(x)=f(y)$ iff there is a bijection between $x$ and $y$. Without the axiom of choice, however, this breaks down in general.