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Let $\kappa$ be an infinite cardinal. Does there always exist a function $f: \kappa \to \kappa$ such that

  1. $f$ is weakly increasing ($f(\alpha) \leq f(\beta)$ for $\alpha \leq \beta < \kappa$)

  2. $f$ is cofinal (equivalently, the image of $f$ is unbounded)

  3. $f$ lies strictly below the diagnoal (i.e. $f(\alpha) < \alpha$ for all $\alpha <\kappa$)

?

I'm willing to relax condition (3) by allowing $f(\alpha) \leq \alpha$ whenenever $\alpha$ has uncountable cofinality.

Also, $f(\alpha)$ need only be defined for sufficiently large $\alpha$, though I'm pretty sure that doesn't make a difference.

For example, this is possible when $\kappa = \aleph_0$, by taking $f(n) = max(n-1,0)$.

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    $\begingroup$ You'll need to relax condition 3 at least somewhat more -- otherwise $f(0)$ cannot exist! $\endgroup$ – Henning Makholm Jan 27 '18 at 1:04
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First, note that such a function does indeed exist if $\kappa$ has cofinality $\omega$, by an easy extension of the $\kappa=\omega$ case.

I now claim that if $cf(\kappa)>\omega$, then no such function exists. To see this, suppose $cf(\kappa)=\lambda>\omega$ and $f$ has the properties above. Fix an increasing cofinal sequence $S=(s_i)_{i\in\lambda}$ in $\kappa$, and chop $\kappa$ into intervals along this sequence: $I_i=[s_i, s_{i+1})$. Now for each $i$ we have $f(s_i)\in I_j$ for some $j<i$ (ignoring a bounded set of values as usual). Let $h:\lambda\rightarrow\lambda$ be defined by $h(i)=j$ iff $f(s_i)\in I_j$. Then $h$ is regressive, hence constant with value $u$ on a stationary (hence cofinal) set for some $u$. But then $f$ itself can't be both weakly increasing and cofinal, since $f$ maps a cofinal set to a bounded set (namely, $I_u$).

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  • $\begingroup$ Wikipedia only states Fodor's lemma for regular cardinals. Is the statement for cardinals of uncountable cofinality the same? $\endgroup$ – tcamps Jan 26 '18 at 23:14
  • $\begingroup$ @tcamps I've added more details. (My now-deleted comments also worked, but were unnecessarily complicated.) $\endgroup$ – Noah Schweber Jan 26 '18 at 23:56
  • $\begingroup$ @tcamps: math.stackexchange.com/questions/56747/… $\endgroup$ – Asaf Karagila Jan 27 '18 at 12:24

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