Find the cardinality of the given set $\{f:\mathbb{Q}\rightarrow\mathbb{Q}|f $is monotonic function$\}$.
My attempt:-I know the cardinality of the set $\{f:\mathbb{Q}\rightarrow\mathbb{Q}|f $is a function$\}$ is $\aleph_{0}^{\aleph_{0}}=2^{\aleph_{0}}=c$. How to find the cardinality of the given set? Please help me.
As you point out, the cardinality is at most $c$. So we only know need to prove that it is also at least $c$. Let $A=\{0,1\}^{\mathbb{N}}$, which is a very convenient set of cardinality $c$. Then we can define an injection $i$ from $A$ to the set you're interested as follows: for $g\in A$ we define $i(g)$ by $i(g)(x)=x$ if $x<0$ or $g(\lfloor x \rfloor)=0$, $i(g)(x)=\lfloor x \rfloor$ if $g(\lfloor x \rfloor)=1$, i.e. $i(g)$ is just the identity for $x<0$ and then on each interval $[n,n+1)$ it's either the identity or the constant function $n$ - then $i(g)$ is monotone and $i$ is an injection so the set of all monotone functions on $\mathbb{Q}$ has cardinality $c$.