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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.

Asaf Karagila
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Paul Slevin
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1 Answers1

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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

$$f_\eta:\kappa\to\kappa:\xi\mapsto\sup\{g_\zeta(\xi)+1:\zeta\le\xi\}\;;$$

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 $g_\zeta$.

Brian M. Scott
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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
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    @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
  • So $| {\xi \mid f_{\eta}(\xi) = g_{\zeta}(\xi) }| < \zeta < \alpha < \kappa^+$. Does this guarantee the set has cardinality $< \kappa$? – Paul Slevin Mar 29 '12 at 21:44
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    @Paul: Yes: $\kappa^+$ is the smallest ordinal of cardinality greater than $\kappa$. – Brian M. Scott Mar 29 '12 at 21:45
  • OK great, I had missed out the part where you chose $\alpha \le \kappa$. – Paul Slevin Mar 29 '12 at 21:59