Finding an injection from $2^{\mathbb{N}} $ to the set of well orderings of $\mathbb{N}$ The question is to show that the cardinality of all well orderings of the natural numbers equals to $2^{\aleph_{0}}$ . So we need two injections for it. One way is easy using the identity map but how do I find an injection from a well known set of cardinality $2^{\aleph_{0}}$ to the set of all well orderings of natural numbers?
I found some answers related to this on other threads but they involved cardinal numbers which I am not really familiar with.
Any help would be greatly appreciated!

HINT: Let's say I take the usual ordering on $\mathbb{N}$ and I re-order two elements - e.g., it now goes $1, 0, 2, 3, 4, . . . $ instead of $0, 1, 2, 3, 4, . . . $. Then the result is still a well-ordering.
Similarly, I can swap infinitely many pairs, as long as the pairs don't overlap! E.g. $1, 0, 3, 2, 5, 4, . . .$ is also a well-ordering (where I swap $2n$ and $2n+1$).
So: break the naturals into infinitely many pairs: $(0,1), (2,3), (4,5), . . .$. How many ways are there to get a new well-ordering on $\mathbb{N}$, just by swapping the elements of some (not necessarily all) of these pairs?

Note that this solution gives you $2^{\aleph_0}$ many well-orderings of $\mathbb{N}$, which are distinct but have the same order-type. You might ask whether we can get $2^{\aleph_0}$ many well-orderings of $\mathbb{N}$ which have distinct order types. Surprisingly, the answer is "we don't know" - this is called the Continuum Hypothesis, and it turns out CH is neither provable nor disprovable from the usual axioms of set theory (ZFC)!
