Infinite cardinal arithmetic

For infinite cardinals, choice gives $\kappa +\kappa =\kappa \cdot \kappa =\kappa$. Exponentiation satisfies $({\kappa }^{\lambda }{)}^{\mu }={\kappa }^{\lambda \mu }$, while a countable sum of cardinals has size the maximum of ${\aleph }_0$ and their supremum.

A compatible transfinite family of bijections can enumerate the square of an ordinal by adjoining a new row and then the remaining new column. Starting with $g_0=\varnothing$, extend at a successor by $g_{\alpha +1}(\alpha ,\beta )={\gamma }_{\alpha }+\beta$ for $\beta \le \alpha$ and $g_{\alpha +1}(\beta ,\alpha )={\gamma }_{\alpha }+(\alpha +1)+\beta$ for $\beta <\alpha$. Thus ${\gamma }_{\alpha +1}={\gamma }_{\alpha }+(\alpha +1)+\alpha$. At limits take compatible unions. The two new domain pieces are disjoint; the column is empty at the initial step $\alpha =0$. Each step enumerates consecutive ordinal intervals, proving bijectivity by transfinite induction. Finite stages have ${\gamma }_n=n^2$. Successor and limit cardinality arguments then give the square equality for well-orderable infinite cardinals without using that equality to justify the construction.

There is a natural bijection between functions $M\to K^L$ and functions $M\times L\to K$, obtained by sending $f$ to $(m,l)\mapsto f(m)(l)$. Therefore

Assuming the axiom of choice, every infinite cardinal satisfiesOne proof takes a least counterexample $\kappa$, identifies cardinals with initial ordinals, and well-orders $\kappa \times \kappa$ first by $\max \{\alpha ,\beta \}$. Every proper initial segment then has cardinality below $\kappa$ by minimality, so the whole order has cardinality at most $\kappa$, contradicting the choice of $\kappa$.

For infinite cardinals $\kappa$ and $\lambda$, the axiom of choice and the square of an infinite cardinal giveThe maximum is a lower bound for both operations, while both are bounded above by respectively $\mu +\mu$ and ${\mu }^2$, where $\mu =\max \{\kappa ,\lambda \}$.

For finitely many infinite sets $X_1,\dots ,X_n$, choose one whose cardinality $\kappa$ is largest. Thenso the union has the same cardinality as that largest member.

Pairwise distinct cardinalities do not prevent a countable family from having a largest member. For example, take $X_1={\aleph }_{\omega }$ and $X_{n+2}={\aleph }_n$ for $n<\omega$, viewing cardinals as initial ordinals. Every later set is a subset of $X_1$, soalthough all the cardinalities are different.

If $\lambda$ is infinite and $2\le \kappa \le \lambda$, thenIndeed,