Cardinal number of a group

Question

I have the following group: $$ A = \{f \in \Bbb N \to \Bbb N \mid \exists S\subseteq \Bbb N, \left.f\right|_S = \operatorname{Id}_S \land f(\Bbb N \setminus S)\subseteq S\} $$

I need to find $|A|$?

Answer 1

$$\begin{array}{l} A &= \{f : \Bbb N \to \Bbb N \mid ∀k \in \Bbb N,f(f(k))=f(k) \}\\ &= \{f : \Bbb N \to \Bbb N \mid \exists S\subseteq \Bbb N, \left.f\right|_S = \operatorname{Id}_S \land f(\Bbb N \setminus S)\subseteq S\}\\ &= \bigcup_{S\subseteq \Bbb N} \{f: \Bbb N \to \Bbb N \mid \left.f\right|_S = \operatorname{Id}_S \land f(\Bbb N \setminus S)\subseteq S\}\\ &\supseteq \bigcup_{S\subseteq \Bbb N} \left\{f:\begin{array}{l}\Bbb N \to \Bbb N \\ x \mapsto\left\{\begin{array}{ll}x&\text{if }x \in S\\\min S&\text{if }x\not\in S\end{array}\right.\end{array}\right\} \end{array}$$

Answer 2

Hint: you could have $f(1)=1, f(2)=2$. You could also have $f(1)=1,f(2)=1$

Answer 3

I think you probably mean the right thing. The set (you do mean set instead of group right?) $A$ contains the set of functions $f_B$ determined by a subset $B$ of $\mathbb{N}\setminus \{0,1\}$ defined by:

$f(0)=0$, $f(1)=1$ and for $x\neq 0,1$, $f(x)=1$ if $x\in B$ and $f(x)=0$ if $x\notin B$. This set of functions is in bijection with $P(\mathbb{N}\setminus\{0,1\})$ which is in bijection with $P(\mathbb{N})$.

Answer 4

the set-endomorphisms of $\mathbb{N}$ do not form a group, but have a monoid structure (with composition as the binary operation).

if $f$ is idempotent then there must be a subset $B_f \subset \mathbb{N}$ such that

(a) $Im(f)$ = $B_f$

(b) $f\mid B_f$ is the identity map on $B $

so for each subset $S \subset \mathbb{N}$ the set of suitable idempotent elements has cardinality $(\mid S \mid)^{\mid \mathbb{N}-S \mid} $

and the number you seek is $\sum_{S \subset \mathbb{N}} (\mid S \mid)^{\mid \mathbb{N}-S \mid} $ = $2^{\aleph_0}$, or thereabouts