TITLE: cardinality of the set of $ \varphi: \mathbb N \to \mathbb N$ such that $\varphi$ is an increasing sequence
QUESTION [8 upvotes]: I know that the set of functions $ f:\mathbb N \to \mathbb N$ is uncountable, but what if we consider only $f$ such that $f$ is increasing? I want to know if this set is countable D: and also the case of bijective and increasing (clearly if the firstone holds then this too). I think that it could be possible that it's countable

REPLY [3 votes]: A nice idea how to prove this was given in this answer by Srivatsan. I am reposting it here since the other question addresses several points, so this might be easily missed among solutions of other parts of that question.

Consider only maps $\varphi \colon \mathbb N\to\mathbb N$ such that $k$ is mapped either to $2k$ or to $2k+1$. 
Every such map is increasing.
We have two possibilities how to choose $\varphi(k)$ for each $k$, so there is $2^{\aleph_0}=\mathfrak c$ such functions. (To make this argument more rigorous, you can show that you can obtain for each element of $\{0,1\}^{\mathbb N}$ a different function of this form.)