- #1

- 177

- 9

**Problem**: Find the cardinality of the set ## A = \{f \in \Bbb N \to \Bbb N. \forall n\leq m .f(n) \geq f (m) \} ##.

I know that ## A \subseteq P(\Bbb N \times \Bbb N) ## implies ## |A| \leq |P(\Bbb N \times \Bbb N)| = | P(\Bbb N) | = \aleph ##. So I have a feeling that ## \aleph \leq |A| ## and this is what I want to show by finding an injective function from a set whose cardinality is ## \aleph ## to set ## A ## and then I'll use Cantor-Schroder-Bernstein in order to deduce that ## |A| = \aleph ##.

**Attempt 1:**

Define ## F : P( \Bbb N ) \to A ## as ## F(\tau)(n) = \sum \limits_{a \in \tau} a ~~~~~~~ , \forall \tau \in P( \Bbb N ), \forall n \in \Bbb N ##

**Attempt 2:**

Define ## F : P( \Bbb N ) \to A ## as ##

F(\tau)(n) = \begin{cases}

\sum \limits_{a \in \tau \setminus\{n\}} a &n\in \tau

\\\sum \limits_{a \in \tau \cup \{n\}} a &n\notin \tau \end{cases} ~~~~~~~ , \forall \tau \in P( \Bbb N ), \forall n \in \Bbb N

##

However, Attempt #1 is bad since it's not injective ## ( F({3,5,7})(n) = F({15})(n) = 15 , \forall n \in \Bbb N)## and I don't even know if attempt #2 is ok or not I tried checking but couldn't get to results. Do you have any ideas as to what the injective function should be? Maybe ## A ## is actually countable?