 Solution

If zero is admitted as a limit case, the assertion there is immediate since ${\gamma }_0=0$. For a nonzero limit, let $\kappa =|\lambda |$, an infinite initial ordinal. There are two cases; separating them avoids choosing an entire family of bijections at earlier stages.

If $\kappa <\lambda$, the hypothesis applies at $\kappa$. Part (a) gives a bijection $g_{\kappa }:{\kappa }^2\to {\gamma }_{\kappa }$, whose range has cardinality $\kappa$ by hypothesis. Fix one bijection $\lambda \to \kappa$; taking its product gives ${\lambda }^2\cong {\kappa }^2$. Since $g_{\lambda }$ is also bijective onto its range, these maps give $\boxed{|{\gamma }_{\lambda }|=\kappa =|\lambda |}$.

If $\lambda =\kappa$ is itself an initial ordinal, suppose ${\gamma }_{\lambda }>\kappa$. Continuity gives ${\gamma }_{\lambda }={\bigcup }_{\alpha <\kappa }{\gamma }_{\alpha }$, so some earlier ${\gamma }_{\alpha }$ contains the ordinal $\kappa$. If $\alpha$ is finite then ${\gamma }_{\alpha }={\alpha }^2$ is finite, impossible. If $\alpha$ is infinite, the hypothesis gives $|{\gamma }_{\alpha }|=|\alpha |<\kappa$, again impossible because its ordinal range contains an initial segment of size $\kappa$.

Thus ${\gamma }_{\lambda }\le \kappa$. Conversely, transfinite induction gives ${\gamma }_{\alpha }\ge \alpha$: each successor appends a nonempty row and limits take unions. In particular ${\gamma }_{\lambda }\ge \lambda =\kappa$. Hence ${\gamma }_{\lambda }=\kappa$ in this case, proving the required cardinal equality as well. The argument includes $\lambda =\omega$, whose earlier stages are finite.