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Logic路issues路in路the路construction路of路the路transfinite路
sequence路of路subcomplexes路$K^\bullet_\alpha$.路Cases路(1)路
and路(3)路are路independent路of路previous路steps,路but路it路seems路
(2)路is路not:路it路requires路to路make路a路choice路since路
$N_\alpha^\bullet$路is路not路unique.路Instead路of路transfinite路
recursion路on路$\alpha$,路maybe路what路one路actually路needs路is路
the路generalized路axiom路of路dependent路choices?路Let路$\kappa$路
be路an路aleph.路This路axiom路says:卢