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The Knuth arrow notation is most often defined only on the natural numbers, but it seems that the central idea of it can be naturally extended to the ordinals, as follows:

$$\alpha\uparrow^0\beta=1$$ $$\alpha\uparrow^1\beta=\alpha^\beta$$ $$\alpha\uparrow^\eta\beta=\sup_{\eta'\lt\eta}[\alpha\uparrow^{\eta'}\sup_{\beta'\lt\beta}(\alpha\uparrow^\eta\beta')].$$

This definition simply replaces the use of $n-1$ and $b-1$ in the usual natural number definition of the Knuth arrow with a supremum over all smaller values, which allows the definition to work with limit ordinals, which have no immediate predecessor. On finite and successor values, this definition agrees with the standard formula. (I'm not sure if others have proposed a different transfinite extension of the concept....I can imagine alternative definitions that treat the limit cases differently, but this won't affect the main conclusion.)

Using this definition, one can show by transfinite induction that if $\alpha, \beta$ and $\eta$ are countable ordinals, then $\alpha\uparrow^\eta\beta$ is also countable, because by the induction hypothesis, this will be a countable supremum of countable ordinals.

In particular, $\omega\uparrow^\omega\omega$ is a very large countable ordinal, and the anwer to your question is that it has cardinality $\aleph_0$.

The Knuth arrow notation is most often defined only on the natural numbers, but the central idea of it can be easily extended to the ordinals, for example as follows:

$$\alpha\uparrow^0\beta=\alpha\beta$$ $$\alpha\uparrow^\eta 0=1\qquad\text{for }\eta\geq 1$$ $$\alpha\uparrow^\eta\beta=\sup_{\eta'\lt\eta}[\alpha\uparrow^{\eta'}\sup_{\beta'\lt\beta}(\alpha\uparrow^\eta\beta')]\qquad\text{otherwise}.$$

This definition simply replaces the use of $n-1$ and $b-1$ in the usual natural number definition of the Knuth arrow with a supremum over all smaller values, which allows the definition to work with limit ordinals, which have no immediate predecessor. On finite and successor values, this definition agrees with the standard formula.

Please note that there are other natural definitions, depending on how one treats the limit stage. which will not be the same as this one. For example, in the definition above, by $\alpha\beta$ I had intended that one should use the natural product, rather than the common product, because this achieves some nicer properties. But others may want to do things differently, and the resulting functions will differ. Nevertheless, what I say below will apply to all the natural formulations of the arrow.

Using this definition, one can show by transfinite induction that if $\alpha, \beta$ and $\eta$ are countable ordinals, then $\alpha\uparrow^\eta\beta$ is also countable, because by the induction hypothesis, this will be a countable supremum of countable ordinals.

In particular, $\omega\uparrow^\omega\omega$ is a very large countable ordinal, and the anwer to your question is that it has cardinality $\aleph_0$.

The Knuth arrow notation is most often defined only on the natural numbers, but it seems that the central idea of it can be naturally extended to the ordinals, as follows:

$$\alpha\uparrow^0\beta=1$$ $$\alpha\uparrow^1\beta=\alpha^\beta$$ $$\alpha\uparrow^\eta\beta=\sup_{\eta'\lt\eta}[\alpha\uparrow^{\eta'}\sup_{\beta'\lt\beta}(\alpha\uparrow^\eta\beta')].$$

This definition simply replaces the use of $n-1$ and $b-1$ in the usual natural number definition of the Knuth arrow with a supremum over all smaller values, which allows the definition to work with limit ordinals, which have no immediate predecessor. On finite and successor values, this definition agrees with the standard formula. (I'm not sure if others have proposed a different transfinite extension of the concept....)

Using this definition, one can show by transfinite induction that if $\alpha, \beta$ and $\eta$ are countable ordinals, then $\alpha\uparrow^\eta\beta$ is also countable, because by the induction hypothesis, this will be a countable supremum of countable ordinals.

In particular, $\omega\uparrow^\omega\omega$ is a very large countable ordinal, and the anwer to your question is that it has cardinality $\aleph_0$.

The Knuth arrow notation is most often defined only on the natural numbers, but it seems that the central idea of it can be naturally extended to the ordinals, as follows:

$$\alpha\uparrow^0\beta=1$$ $$\alpha\uparrow^1\beta=\alpha^\beta$$ $$\alpha\uparrow^\eta\beta=\sup_{\eta'\lt\eta}[\alpha\uparrow^{\eta'}\sup_{\beta'\lt\beta}(\alpha\uparrow^\eta\beta')].$$

This definition simply replaces the use of $n-1$ and $b-1$ in the usual natural number definition of the Knuth arrow with a supremum over all smaller values, which allows the definition to work with limit ordinals, which have no immediate predecessor. On finite and successor values, this definition agrees with the standard formula. (I'm not sure if others have proposed a different transfinite extension of the concept....I can imagine alternative definitions that treat the limit cases differently, but this won't affect the main conclusion.)

Using this definition, one can show by transfinite induction that if $\alpha, \beta$ and $\eta$ are countable ordinals, then $\alpha\uparrow^\eta\beta$ is also countable, because by the induction hypothesis, this will be a countable supremum of countable ordinals.

In particular, $\omega\uparrow^\omega\omega$ is a very large countable ordinal, and the anwer to your question is that it has cardinality $\aleph_0$.

The Knuth arrow notation is most often defined only on the natural numbers, but it seems that the central idea of it can be naturally extended to the ordinals, as follows:

$$\alpha\uparrow^0\beta=1$$ $$\alpha\uparrow^1\beta=\alpha^\beta$$ $$\alpha\uparrow^\eta\beta=\sup_{\eta'\lt\eta,\ \beta'\lt\beta}[\alpha\uparrow^{\eta'}(\alpha\uparrow^\eta\beta')].$$

This definition simply replaces the use of $n-1$ and $b-1$ in the usual natural number definition of the Knuth arrow with a supremum over all smaller values, which allows the definition to work with limit ordinals, which have no immediate predecessor.

Using this definition, one can show by transfinite induction that if $\alpha, \beta$ and $\eta$ are countable ordinals, then $\alpha\uparrow^\eta\beta$ is also countable, because by the induction hypothesis, this will be a countable supremum of countable ordinals.

In particular, $\omega\uparrow^\omega\omega$ is a very large countable ordinal, and the anwer to your question is that it has cardinality $\aleph_0$.

The Knuth arrow notation is most often defined only on the natural numbers, but it seems that the central idea of it can be naturally extended to the ordinals, as follows:

$$\alpha\uparrow^0\beta=1$$ $$\alpha\uparrow^1\beta=\alpha^\beta$$ $$\alpha\uparrow^\eta\beta=\sup_{\eta'\lt\eta}[\alpha\uparrow^{\eta'}\sup_{\beta'\lt\beta}(\alpha\uparrow^\eta\beta')].$$

This definition simply replaces the use of $n-1$ and $b-1$ in the usual natural number definition of the Knuth arrow with a supremum over all smaller values, which allows the definition to work with limit ordinals, which have no immediate predecessor. On finite and successor values, this definition agrees with the standard formula. (I'm not sure if others have proposed a different transfinite extension of the concept....)

Using this definition, one can show by transfinite induction that if $\alpha, \beta$ and $\eta$ are countable ordinals, then $\alpha\uparrow^\eta\beta$ is also countable, because by the induction hypothesis, this will be a countable supremum of countable ordinals.

In particular, $\omega\uparrow^\omega\omega$ is a very large countable ordinal, and the anwer to your question is that it has cardinality $\aleph_0$.