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Let $\kappa$ be a regular cardinal. I want to show that there exist $\kappa^+$ almost disjoint functions $\kappa \to \kappa$. Recall that this means for any two such functions $f,g$ we have $| \{\alpha \mid f(\alpha) = g(\alpha) \}| < \kappa$.

Apparently it is enough to show that given $\kappa$ almost disjoint functions $\{ f_\nu \mid \nu < \kappa\}$, then there exists an $f:\kappa \to \kappa$ almost disjoint from each $f_\nu$, $(\nu < \kappa)$. I can't see why this is sufficient to prove the statement though - can anybody enlighten me? Thanks very much.

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I think we can use induction to construct $\kappa^+$ functions given we have such a family $\{ f_\nu \mid \nu < \kappa \}$ but I'm not sure how to initially construct such a family. – Paul Slevin Mar 29 '12 at 20:47
There’s nothing special needed initially. – Brian M. Scott Mar 29 '12 at 20:52
up vote 5 down vote accepted

Just construct $\{f_\xi:\xi<\kappa^+\}$ recursively. At each stage you have at most $\kappa$ functions already defined, so by your hypothesis you can add a new one that’s almost disjoint from the ones that you already have.

In other words, at stage $\eta<\kappa^+$ you have $\mathscr{F}_\eta=\{f_\xi:\xi<\eta\}$. $|\mathscr{F}_\eta|\le\kappa$, so by your hypothesis there is a function $f_\eta:\kappa\to\kappa$ that is almost disjoint from $f_\xi$ for each $\xi<\eta$. Now continue.

Added: For the sake of completeness I’m adding the construction of $f_\eta$. Since $|\mathscr{F}_\eta|\le\kappa$, we can re-index it as $\{g_\xi:\xi<\alpha\}$ for some $\alpha\le\kappa$. Now define


then for each $\zeta<\alpha$ we have $f_\eta(\xi)>g_\zeta(\xi)$ whenever $\xi\ge\zeta$, and hence $f_\eta$ is almost disjoint from $g_\zeta$.

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Thank you. It says here I can use a diagonal argument to construct such functions, but I'm having trouble parsing the text. Is it possible we can define $f_{\eta}$ such that $f_{\eta}(\beta) \not= f_{\xi}(\beta)$ for all $\xi < \eta$ and $\beta > \xi$ ? – Paul Slevin Mar 29 '12 at 21:10
@Paul: My construction of $f_\eta$ in the added bit is the diagonal argument in question. – Brian M. Scott Mar 29 '12 at 21:14
Oops I was being a bit dense. I think a suitable function is given any $\beta < \kappa$, we ensure that $f_\xi (\beta) \not= f_\alpha (\beta)$ whenever $\beta > \xi$. – Paul Slevin Mar 29 '12 at 21:16
Ah OK, I didn't realise you had added a bit. Thanks very much. – Paul Slevin Mar 29 '12 at 21:16
@Paul: Yes: $\kappa^+$ is the smallest ordinal of cardinality greater than $\kappa$. – Brian M. Scott Mar 29 '12 at 21:45

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