 Solution

The finite-stage calculation gives ${\gamma }_{\omega }={\sup }_{n<\omega }{n}^2=\omega$, so the desired equality holds at the first infinite ordinal. Apply transfinite induction to all subsequent infinite ordinals. The successor step is part (b), and the limit step is part (c). ThusBy the well-ordering theorem, any infinite cardinal in the question may be represented by an infinite initial ordinal $\kappa$. The corrected recursion supplies a bijection $g_{\kappa }:\kappa \times \kappa \to {\gamma }_{\kappa }$, and the just-proved equality gives $|{\gamma }_{\kappa }|=\kappa$. ThereforeThe proof used only finite-copy interleaving before reaching this conclusion; using the square equality itself to estimate the successor or limit stages would have been circular.