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Can there be a set of functions $S\subseteq\{f \mid f:\mathbb{N}\to\mathbb{N}\}$ of cardinality $\mathfrak{c}$ (real numbers) such that the set $S_f:=\{g \in S \mid g(n)\leq f(n), \forall n \in \mathbb{N}\}$ is finite for all $f:\mathbb{N}\to \mathbb{N}$?

My intuition says no, since there have to be uncountably many $f$s in $S$ that take a certain value at each $n\in\mathbb{N}$. How would you prove this?

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    $\begingroup$ Do you mean $S_f$ is finite for all $f$ or do you mean $S_f$ is finite for all $f\in S$? $\endgroup$
    – user1318062
    Commented May 5 at 1:02
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    $\begingroup$ If you mean for all $f$ then the answer is no by the diagonal argument, if you mean for all $f\in S$ the answer is yes. $\endgroup$
    – user1318062
    Commented May 5 at 1:21

2 Answers 2

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Assume $S$ is uncountable.

Then there is some $N\in \mathbb{N}$ such that $g(0) \leq N$ for uncountably many $g\in S$. Let $f(0)$ be the smallest such $N$, define $S_0$ to be the set of all $g \in S$ such that $g(0) \leq f(0)$, and choose $g_0 \in S_0$.

Inductively perform the following:

  • Note that there is some $N$ such that $g(n) \leq N$ for uncountably many $g\in S_{n-1}$. Let $N_n$ be the smallest such $N$ and define $$f(n) = N_n + \sum_{k=0}^{n-1}g_k(n)$$
  • Define $S_n$ to be the set of all $g \in S_{n-1}$ such that $g(n) \leq f(n)$. Note that $S_n$ is uncountable and $g_k \in S_n$ for every $k < n$.
  • Choose $g_n \in S_n$ such that $g_n \neq g_k$ for any $k < n$.

For such an $f$, $S_f \supseteq \{g_k : k \in \mathbb{N}\}$, so $S_f$ is not finite.


I initially read the condition on $S_f$ as $S_f$ is finite for each $f \in S$. With that [mis]reading, we have the following example:

For each subset $A \subseteq \mathbb{N}$ define the function $f_A$ by $$k \in A \implies f(2k) = 0, f(2k+1) = 2 \\ k \not\in A \implies f(2k)=f(2k+1) = 1$$

Then $S = \{f_A : A \subseteq \mathbb{N}\}$ is an antichain (i.e. $|S_f|=1$ for each $f \in S$) which is equipotent with $\mathcal{P}(\mathbb{N})$, which is uncountable.

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    $\begingroup$ I had the same idea but instead I had $f$ the indicator function of $A$ and $g$ the indicator function of $A^c$ and then put them together. But as it stands $f$ is unrestricted so there is no such $S$. $\endgroup$
    – user1318062
    Commented May 5 at 1:27
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For $N\subset\mathbb{N}$ let $$f_N(n+1)=\mathbb{1}_{n\in N}$$ where $\mathbb{1}_{n\in N}$ is $1$ if $n\in N$ and $0$ else and let $$f_N(0)=-\vert N \vert.$$ Now let $$S=\{f_N:N\subset\mathbb{N}\}.$$ Clearly for $N,M\subset\mathbb{N}$ with $N\neq M$ we have $$f_N\not\le f_M\text{ and } f_M\not\le f_N$$ so $$\vert S_{f_N}\vert=1$$ for all $N\subset\mathbb{N}$.

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