The question: Decide with proof which has greater Cardinality $\Bbb{N}^\Bbb{N}$ or $2^\Bbb{N}$.
My intuition: They will be the same. By Cantors argument and the continuum hypothesis, both will have cardinality $\aleph_1$.
My attempt at a proof: I know that $|A|\le|B|$ if there is an injection $f:A \rightarrow B$ and by the Schroder Bernstein theorem I know that you can always compare cardinalities. Thus, I want to show that there is an injection from $\Bbb{N}^\Bbb{N} \rightarrow 2^\Bbb{N}$ and one from $2^\Bbb{N} \rightarrow \Bbb{N}^\Bbb{N}$.
I thought an injection from $2^\Bbb{N} \rightarrow \Bbb{N}^\Bbb{N}$ might be: for each element of the power set of Naturals, arrange its elements from smallest to largest and map the set to the element of $\Bbb{N}^\Bbb{N}$ that has all of the same elements with an infinite string of zeros if needed. For example: ${(a_1,a_2,...,a_k)} \in 2^\Bbb{N} \rightarrow {(a_1, a_2, a_3,...,a_k,0,0,0,0...)}\in \Bbb{N}^\Bbb{N}$ where $a_1<a_2<a_3<...$
Is this a valid injection? Can you please help me find an injection the other way or am I not on the right track? Thanks.