Suppose $\alpha,\beta$ are ordinals with cardinality $\kappa$. It can be proved by induction that if $\kappa=\aleph_0$, then $\alpha+\beta,\alpha\cdot\beta,\alpha^\beta$ also have cardinality $\kappa$.
But may I please ask：in general, for which $\kappa$ can we have $\#\alpha=\#\beta=\kappa$ implies the following:
(1)$\alpha+\beta$
(2)$\alpha\cdot\beta$
(3)$\alpha^\beta$
Has cardinality $\kappa$?
So far, I think that for any finite $\kappa\ne 0$, we cannot have (1)(2)(3) has the same cardinality $\kappa$. And for $\kappa=0$, the cardinality of (1)(2)(3) are all zero. And I think for any infinite $\kappa$, (1)(2)(3) has cardinality $\kappa$. Is that correct? Or if something is wrong, could someone give a correct argument with proof please? Thanks in advance!
EDIT: So is that the fact that: When $\kappa=0$, the cardinality of (1)(2) are zero and thus has cardinality $\kappa$, (3) has cardinality $1$, so its cardinality is not $\kappa$.
When $\kappa\ne 0$, $\kappa$ is finite, the cardinality of (1)(2)(3) can never be $\kappa$.
When $\kappa$ is infinite, (1)(2)(3) has cardinality $\kappa$.
Could someone please confirm if these are fine now? Thanks in advance!
If $\kappa$ is infinite, then it is true that $\alpha+\beta$, $\alpha\cdot\beta$, and $\alpha^\beta$ all have cardinality $\kappa$. How you prove this depends on what definitions of ordinal arithmetic you have. It is simplest to prove if you have explicit definitions of what the order types are: $\alpha+\beta$ is the disjoint union of $\alpha$ and $\beta$ ordered such that $\beta$ comes after $\alpha$, $\alpha\cdot\beta$ is the Cartesian product $\beta\times\alpha$ with the lexicographic order, and $\alpha^\beta$ is the set of finite support functions $\beta\to\alpha$ with the lexicographic order with respect to the reverse order on $\beta$. With these definitions, it is clear that $$|\alpha+\beta|=\kappa+\kappa=\kappa$$ and $$|\alpha\cdot\beta|=\kappa\cdot\kappa=\kappa$$ (where the operations on the right are cardinal arithmetic). For the exponential, note that there are only $\kappa$ many finite subsets of $\beta$ and only $\kappa$ functions from any finite set to $\alpha$, so $$|\alpha^\beta|\leq\kappa\cdot\kappa=\kappa.$$ On the other hand, clearly $|\alpha^\beta|\geq\kappa$, so $|\alpha^\beta|=\kappa$.
The finite case is easier, since then the ordinal arithmetic operations just coincide with the usual natural number operations and ordinals and cardinals are the same. So if $\kappa>0$, $\alpha+\beta=\kappa+\kappa>\kappa$. If $\kappa>1$, $\alpha\cdot\beta=\kappa\cdot\kappa>\kappa$ and $\kappa^\kappa>\kappa$. On the other hand, for $\kappa=1$, we have $1\cdot 1=1$ and $1^1=1$. And for $\kappa=0$, we have $0+0=0$ and $0\cdot 0=0$ (but $0^0=1$). So there are no finite $\kappa$ such that all three of your conditions hold, although two of them hold for $\kappa=0$ and $\kappa=1$.