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Jan 29, 2011 at 20:34 comment added David Feldman Thanks both, right, I garbled my own question. Think I've improved it now.
Jan 29, 2011 at 20:21 history edited David Feldman CC BY-SA 2.5
edited title; added 323 characters in body
Jan 29, 2011 at 16:21 history edited Andrés E. Caicedo CC BY-SA 2.5
added 6 characters in body
Jan 29, 2011 at 9:47 answer added Not Mike timeline score: 2
Jan 29, 2011 at 9:17 comment added Not Mike For any normal space $X$ we have $\\{\\{x\\}: x\in X\\} \subset H(1)$ so $\bigcup_{\lambda\le1} H(\lambda)$ must contain $\\{\\{x\\}: x\in X\\}$ and every point of $X$. So the hierarchy is degenerate, and I don't think what you intended.
Jan 29, 2011 at 9:14 comment added Joel David Hamkins Won't $X$ itself be in every family, since it is the zero set of the constant zero function? And in this case, we add it back in at each step. Could you explain? Also, in (1) you haven't described $H(0)$ as a family of sets, and in (2) you are taking the union of families, which would be a family, but I think you intend to take the union of the sets in that union family---could you clarify?
Jan 29, 2011 at 7:11 history asked David Feldman CC BY-SA 2.5