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Let $\kappa$ be an uncountable regular cardinal and let $[\kappa]^{\aleph_0}$ be the set of all countably infinite subsets of $\kappa$. A (proper) ideal $I$ on $[\kappa]^{\aleph_0}$ is then a non-empty collection of subsets of $[\kappa]^{\aleph_0}$ such that

  1. $[\kappa]^{\aleph_0}\notin I$
  2. If $X\in I$ and $Y\subseteq X$ then $Y\in I$
  3. If $X,Y\in I$ then $X\cup Y\in I$.

My question is then what normality of $I$ means, as I've stumbled across two ostensibly different definitions. Recall that a set $X\subseteq [\kappa]^{\aleph_0}$ is $I$-positive if $X\notin I$.

Definition 1. $I$ is normal if whenever $X$ is $I$-positive and $f:X\to\kappa$ is a function satisfying that $f(\sigma)\in\sigma$ then there's an $I$-positive $Y\subseteq X$ on which $f$ is constant.

Definition 2. $I$ is normal if whenever $X$ is $I$-positive and $f:X\to\kappa$ is a function satisfying that $f(\sigma)<\sup(\sigma)$ for all $\sigma\in X$, then there's an $I$-positive $Y\subseteq X$ on which $f$ is constant.

So it boils down to what a regressive function in this context is. Are these definitions equivalent? For the sake of argument we can also assume that $I$ is fine, meaning that $\{\sigma\in[\kappa]^{\aleph_0}\mid\xi\notin\sigma\}\in I$ for every $\xi<\kappa$, and countably complete, meaning that it's closed under countable unions.

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  • $\begingroup$ What is $\sigma(i)$ here? The $i$th member in some fixed enumeration, or...? $\endgroup$ – Asaf Karagila Mar 12 '18 at 11:06
  • $\begingroup$ Ah right. Yeah the $i$'th member in increasing order. $\endgroup$ – Dan Saattrup Nielsen Mar 12 '18 at 11:18
  • $\begingroup$ But that doesn't make sense, as that would only allow $\omega$-type subsets, and there are far more countable sets. $\endgroup$ – Asaf Karagila Mar 12 '18 at 11:22
  • $\begingroup$ Hm, you're right. I suppose we could fix some enumeration, but in that case the definition suddenly depends upon the choice of enumeration. Instead, I'll just remove that second definition. $\endgroup$ – Dan Saattrup Nielsen Mar 12 '18 at 11:27
  • $\begingroup$ It should be enough to prove that a positive set has limit order type. From there it should be fairly simple. $\endgroup$ – Asaf Karagila Mar 12 '18 at 13:14
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Here's an answer. The argument for the 'only if' direction is due to Stamatis Dimopoulos.

Claim. Let $I$ be a fine ideal on $[\kappa]^{\aleph_0}$ satisfying the first normality definition. Then $I$ satisfies the second normality definition iff it's $\kappa$-complete.

Proof. $(\Rightarrow)$ Assume that $I$ satisfies the second normality defintion. Let $\gamma<\kappa$ and $\langle X_\xi\mid\xi<\gamma\rangle\in{^\gamma}I$ be a $\gamma$-sequence of elements of $I$. Note that the second definition implies that $I$ is closed under the following ostensibly stronger notion of diagonal union:

$$ \mathscr{D}_{\xi<\kappa}Z_\xi:=\{\sigma\in[\kappa]^{\aleph_0}\mid\sigma\in\bigcup_{\xi<\sup(\sigma)}Z_\xi\}$$

Define $Z_\xi:=X_\xi$ if $\xi<\gamma$ and $Z_\xi:=\emptyset$ for $\xi\in[\gamma,\kappa)$, so that $\mathscr{D}\vec Z\in I$. If now $\sigma\in X_\xi$ for some $\xi<\gamma$ then either $\sup(\sigma)<\gamma$ or $\sigma\in\mathscr{D}\vec Z$. Since

$$Y:=\{\sigma\in[\kappa]^{\aleph_0}\mid\sup(\sigma)<\gamma\}=\bigcap_{\xi<\gamma}\{\sigma\in[\kappa]^{\aleph_0}\mid\xi\notin\gamma\}\in I$$

by fineness, this means that $\bigcup\vec X\subseteq\mathscr{D}\vec Z\cup Y\in I$ and so $\bigcup\vec X\in I$, making $I$ $\kappa$-complete.

$(\Leftarrow)$ Assume that $I$ is $\kappa$-complete, let $X$ be $I$-positive and let $f:X\to\kappa$ satisfy $f(\sigma)<\sup(\sigma)$. Define $g:X\to\kappa$ as $g(\sigma)$ being the least $\xi\in\sigma$ such that $f(\sigma)<\xi$. Then $g(\sigma)\in\sigma$ for all $\sigma\in X$, so by definition 1 we get an $I$-positive $Y\subseteq X$ with $g[Y]=\{\xi_0\}$ for some $\xi_0<\kappa$. Define now for each $\eta<\xi_0$ the set $Y_\eta:=\{\sigma\in Y\mid f(\sigma)=\eta\}$, and note that $Y=\bigcup_{\eta<\xi_0}Y_\eta$. Then $\kappa$-completeness of $I$ implies that there's some $\eta_0<\xi_0$ such that $Y_{\eta_0}$ is $I$-positive. As every $\sigma\in Y_{\eta_0}$ satisfies that $f(\sigma)=\eta_0$ we get that $I$ satisfies definition 2. QED

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  • $\begingroup$ Tell Stamatis that dancing on two weddings always ends up with the dancer falling down and breaking their hip. $\endgroup$ – Asaf Karagila Mar 14 '18 at 12:14

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