Find cardinality of $$B = \left\{ f : \mathbb N \rightarrow \mathbb N \mid \forall n(f(n)\le n) \wedge \forall m\; \exists n (f(n) > m) \right\}$$

My try

I have solved this, but I am not sure if it is correct (I have not a lot of experience in set theory). Can somebody check that or give some tips (or both)?
$|B| < \mathfrak{c}$ because $|B| < |\mathbb N|^{|\mathbb N|}$
from the other hand I can define $G$ injective such as:
$$G: (\mathbb N \rightarrow \left\{0,1 \right\}) \rightarrow B $$ $$ G(\alpha)(n) = \begin{cases} \alpha(n) + G(\alpha)(n-1), &\text{if }n \neq 0 \\ 0, &\text{if }n = 0. \end{cases} $$

The function $G$ increases and is injective, and its power is $\mathfrak{c} $ so $|B| = \mathfrak{c}$

  • 1
    $\begingroup$ Note that if $\alpha$ takes the value $1$ only finitely many times, then $G(\alpha) \notin B$. $\endgroup$ – Mees de Vries Jan 30 at 16:26
  • 1
    $\begingroup$ Why do you say $|B| < |\mathbb{N}|^{|\mathbb{N}|}$? $\endgroup$ – Clive Newstead Jan 30 at 17:06
  • $\begingroup$ It's stil not clear what is $\mathfrak{c}$, it should be defined when first used. Which power of $G$ ? $\endgroup$ – Soleil Jan 30 at 18:21
  • $\begingroup$ continuum - why is it not clear? $\endgroup$ – VirtualUser Jan 30 at 18:35
  • $\begingroup$ @VirtualUser Because it s not defined in usual set theory / decriptive set theory books, and you did not defined it. Hence $\mathfrak{c}:= 2^{\mathbb N} = \aleph_1$. $\endgroup$ – Soleil Jan 30 at 18:48

$B$ is the set of functions $\mathbb{N} \to \mathbb{N}$ that are unbounded in absolute terms, but are bounded by the identity function.

Evidently $|B| \le \mathfrak{c}$ since $B$ is a subset of $\mathbb{N}^{\mathbb{N}}$.

To see that $\mathfrak{c} \le |B|$, let $\mathcal{P}_{\mathsf{inf}}(\mathbb{N}) \subseteq \mathcal{P}(\mathbb{N})$ be the set of infinite subsets of $\mathbb{N}$. Then $|\mathcal{P}_{\mathsf{inf}}(\mathbb{N})| = \mathfrak{c}$, since the set of all finite subsets of $\mathbb{\mathbb{N}}$ is countable.

For each $U \in \mathcal{P}_{\mathsf{inf}}(\mathbb{N})$, define $f_U \in B$ inductively by $$f_U(0) = 0 \quad \text{and} \quad f_U(n+1) = \begin{cases} f_U(n) & \text{if } n \not\in U \\ f_U(n)+1 & \text{if } n \in U \end{cases}$$

The fact that $f_U$ is unbounded follows from the fact that $U$ is infinite. You can prove by induction that $f_U(n) \le n$ for all $n \in \mathbb{N}$, and that $U \mapsto f_U$ defines an injection $\mathcal{P}_{\mathsf{inf}}(\mathbb{N}) \to B$.

Hence $\mathfrak{c} \le |B|$.


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