Are these definitions of normality equivalent? 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

*

*$[\kappa]^{\aleph_0}\notin I$

*If $X\in I$ and $Y\subseteq X$ then $Y\in I$

*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.
 A: 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

