Proof Check: The cardinality of all monotonically increasing series of natural numbers. Given an infinite series, $a_1a_2a_3\ldots$ define $F$, $F(a_1a_2a_3\ldots) = a_1\bmod(2)a_2\bmod(2)a_3\bmod(2)\ldots$
It's trivial to show that $F$ is onto the set of all infinite binary series, which has a cardinality of $\aleph_1$. 
So the cardinality of the aforementioned set $\ge \aleph_1$.
Then again the set of all infinite binary sets is a subset of the set of all infinite series of natural numbers, therefore yielding a cardinality of $\le\aleph_1$.
All in all we get a cardinality of $\aleph_1$ by the Cantor–Bernstein–Schroeder theorem.
I feel like I might've made a mistake with $F$, but I'm not sure. Can anyone verify? Thanks for your time!
 A: Your gravest mistake is that $\aleph_1$ is NOT the cardinality of the continuum (well, at least not in general). We define $\aleph_1$ to be the least uncountable cardinal, whereas $2^{\aleph_0}$ is just uncountable. It is consistent with the axioms of set theory that they are equal, and it is consistent that they are not.
The $F$ that you define is fine, and it is indeed a surjection from the set of monotonically increasing sequences onto the set of binary sequences. This establishes, as you point out that the cardinality of the set that you are interested in is at least $2^{\aleph_0}$.
However the second part is not well-written. You provide a second argument as to why the cardinality is at least $2^{\aleph_0}$ rather than providing an argument that it is at most $2^{\aleph_0}$ (which will then imply equality).
HINT: Every sequence of natural numbers is in fact a function $f\colon\Bbb{N\to N}$. Note that such function $f$ is a subset of $\Bbb{N\times N}$. How many subsets does this product have?
