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Construction of $V$ may be described by the following sequences:

  • $\aleph_o$, $\aleph_1$, $\aleph_2$, ..., $\aleph_\omega$;

  • $\aleph_{\omega+1}$,...,$\aleph_{\omega+\omega}$;

  • $\aleph_{\omega+\omega}$, $\aleph_{\omega+\omega+\omega}$, ..., $\aleph_{\sum_{i=0}^{\infty}(\omega_i)}$.

I repeat the subscript of the last $\aleph$ for readability: ${\sum_{i=0}^{\infty}(\omega_i)}$.

Can I define $\aleph_A$ , where $A={\sum_{i=0}^{\infty}}(\omega_i)$?

Herman
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  • What is "$\infty$" here? Certainly there is no largest cardinal (consider e.g. $\mathcal{P}(\kappa)$ for a given cardinal $\kappa$). Also, I think your $w$s should be $\omega$s. – Noah Schweber Mar 28 '21 at 23:02
  • Thank you for the correction. It was meant to express an infinite sum of $\omega_i$. – Herman Mar 28 '21 at 23:08
  • "An infinite sum" What infinite sum exactly? $\sum_{i=0}^\omega\omega_i$ is different from $\sum_{i=0}^{\omega_{17}}\omega_i$ is different from ... The notation "$\infty$" is vague, and makes this impossible to answer. – Noah Schweber Mar 28 '21 at 23:09
  • Thank you! Should I delete the question? – Herman Mar 28 '21 at 23:17

1 Answers1

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The superscript "$\infty$" here is very problematic. The short version is that we can make sense of "$\sum_{i=0}^\lambda \omega_i$" (and consequently $\omega_{\sum_{i=0}^\lambda \omega_i}$) for any ordinal $\lambda$; however, this does not lead to a greatest cardinal, and in particular we cannot sum over all ordinals.

The key to making sense of such big ordinals is the axiom (scheme) of replacement. Intuitively, if we think about the construction of $V$ as an ongoing process the axiom of replacement lets us "bundle together" all the ordinals we've made so far to make even bigger ones. For example, applying replacement to (the definition of) the class function $i\mapsto \omega_i$ we get that for any ordinal $\lambda$ the set $\{\omega_i:i<\lambda\}$ exists, and from this set it's easy to build $\sum_{i=0}^\lambda\omega_i$. Without replacemenet incidentally we can't even build the ordinal $\omega+\omega$. (Less mysteriously, $V_{\omega+\omega}$ satisfies $\mathsf{ZC}$ = $\mathsf{ZFC}$ without replacement, and obviously $\omega+\omega\not\in V_{\omega+\omega}$.)

Noah Schweber
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