${\mathrm{\aleph }}_0\oplus {\mathrm{\aleph }}_0={\mathrm{\aleph }}_0\otimes {\mathrm{\aleph }}_0={\mathrm{\aleph }}_0$, ${\mathrm{\aleph }}_1\oplus {\mathrm{\aleph }}_0={\mathrm{\aleph }}_1$ and $5\oplus {\mathrm{\aleph }}_0={\mathrm{\aleph }}_0$, computed from absorption and, in the countable cases, independently from the published bijection $\omega \times \omega \approx \omega$

Work in ZF; no choice principle is used anywhere below. With $\oplus$ and $\otimes$ as in Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation ${\kappa }^{\lambda }$, and why they are written apart from the ordinal operations and the alephs as in The successor cardinal ${\kappa }^{+}$, the alephs ${\mathrm{\aleph }}_{\alpha }$, the beths ${\mathrm{\beth }}_{\alpha }$, successor and limit cardinals, and the identifications ${\mathrm{\aleph }}_0=\omega$ and ${\mathrm{\aleph }}_1={\omega }_1$:

${\mathrm{\aleph }}_0\oplus {\mathrm{\aleph }}_0={\mathrm{\aleph }}_0,{\mathrm{\aleph }}_0\otimes {\mathrm{\aleph }}_0={\mathrm{\aleph }}_0,{\mathrm{\aleph }}_1\oplus {\mathrm{\aleph }}_0={\mathrm{\aleph }}_1,5\oplus {\mathrm{\aleph }}_0={\mathrm{\aleph }}_0\mathrm{.}$

Each is an instance of absorption (Absorption: for cardinals $\kappa ,\lambda$ with $\kappa$ infinite and $\lambda \le \kappa$, $\kappa \oplus \lambda =\kappa$, and $\kappa \otimes \lambda =\kappa$ when $\lambda \mathrm{\ne }0$). The countable ones are also obtained a second way, from a bijection that was available before cardinal arithmetic existed: $\omega \times \omega \approx \omega$ ($\mathbb{N}\times \mathbb{N}\approx \mathbb{N}$), together with two inclusions and no further input.

What absorption replaces. Step 2.2 is what a reader would have had to do before Absorption: for cardinals $\kappa ,\lambda$ with $\kappa$ infinite and $\lambda \le \kappa$, $\kappa \oplus \lambda =\kappa$, and $\kappa \otimes \lambda =\kappa$ when $\lambda \mathrm{\ne }0$ existed: produce a bijection or a pair of injections for each computation separately. Step 2.1 does all four in one line, and the content of the corollary is exactly that the bookkeeping is unnecessary.

Why ${\mathrm{\aleph }}_1\oplus {\mathrm{\aleph }}_0={\mathrm{\aleph }}_1$ has no second computation here. The countable cases are witnessed by explicit maps because $\omega$ is concrete. At ${\mathrm{\aleph }}_1$ no comparable explicit bijection is written down here; the computation goes through Hessenberg: $\kappa \otimes \kappa =\kappa$ for every infinite cardinal $\kappa$, proved in ZF from the canonical well-order of $\kappa \times \kappa$, which Absorption: for cardinals $\kappa ,\lambda$ with $\kappa$ infinite and $\lambda \le \kappa$, $\kappa \oplus \lambda =\kappa$, and $\kappa \otimes \lambda =\kappa$ when $\lambda \mathrm{\ne }0$ packages. That is not a gap in this example but the reason the general theorem is worth proving.

The finite summand does not vanish for a trivial reason. $5\bigsqcup \omega$ does have more elements than $\omega$ in the naive sense: it carries a tagged copy of $\omega$ and five further points. The equality $5\oplus {\mathrm{\aleph }}_0={\mathrm{\aleph }}_0$ says only that the two sets are equinumerous, and what makes that true is that an infinite well-ordered set absorbs finitely many extra points, the same shift that makes an infinite cardinal a limit ordinal in Every natural number and $\omega$ are cardinals, every infinite cardinal is a limit ordinal, and on the natural numbers the cardinal operations are the published finite counting operations, with $\mid A\mid$ in the finite sense equal to $\mid A\mid$ in the cardinal sense.

Order does not matter here, and that is not automatic. $\oplus$ is commutative, so $5\oplus {\mathrm{\aleph }}_0$ and ${\mathrm{\aleph }}_0\oplus 5$ are the same cardinal. The ordinal $+$ on the very same objects is not commutative, which is precisely why Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation ${\kappa }^{\lambda }$, and why they are written apart from the ordinal operations gives the cardinal operation its own symbol rather than reusing $+$.