I'm reading books on set theory and I came up with the following 'proof' that $\text{card}(\mathcal{P}(\mathbb{N}))=\aleph_1$. What is wrong with it? I really cannot tell!
By $\text{card}(\mathcal{P}(\mathbb{N}))$ I mean the cardinality of the power set of the set of nonnegative integers (I also see $\mathbb{N}$ as the set of finite von Neumann ordinals), which is $2^{\aleph_0}$, and by $\aleph_1$ I mean the cardinality of $\omega_1$, which is the least uncountable ordinal and also the least upper bound of the set of all ordinals of cardinality at most $\aleph_0$. Now, $2^{\aleph_0}=\aleph_1$ is the continuum hypothesis, which is independent of ZFC.)
Proof.
Each $\alpha\in\omega_1$ has cardinality at most $\aleph_0$ and each such $\alpha$ is wellordered by $\in$.
Each $a\in\mathcal{P}(\mathbb{N})$ has cardinality at most $\aleph_0$, and each such $a$ is wellordered by $\in$. This is because every $a\in\mathcal{P}(\mathbb{N})$ is a (possibly infinite) set of (finite) ordinals, and every nonempty set of ordinals is wellordered by $\in$.
We can establish an obvious bijection from $\mathcal{P}(\mathbb{N})$ to some set of ordinals $N$, namely $a\mapsto\text{ord}(a)$, i.e. between $a\in\mathcal{P}(\mathbb{N})$ and the order-type of wellordered set $a$. Each ordinal $\text{ord}(a)\in N$ also has cardinality at most $\aleph_0$, and since we know that $\omega_1$ is the least upper bound of all ordinals of cardinality at most $\aleph_0$ (where $\text{card}(\omega_1)=\aleph_1$), then $\text{card}(N)\leq\aleph_1$.
But $\text{card}(\mathcal{P}(\mathbb{N})) = \text{card}(N)$ (due to the existence of a bijection from $\mathcal{P}(\mathbb{N})$ to $N$), so by a diagonal argument $\text{card}(N)>\aleph_0$. It follows that $\text{card}(N)=\aleph_1$, and $\text{card}(\mathcal{P}(\mathbb{N}))=\aleph_1$.
Note that every set in $\mathcal P(\Bbb N)$ is a subset of $\Bbb N$, so its order type is $\leq\omega$. So this is certainly not a surjection onto $\omega_1$.
You can define a surjection onto $\omega_1$ by considering each set of natural numbers encoding a set of pairs of natural numbers. But then you will also face things which are not well-orders (e.g. things isomorphic to $\Bbb Z$ or $\Bbb Q$ and things which are not partial orders to begin with). And not to mention that there are $2^{\aleph_0}$ subsets of $\Bbb{N\times N}$ which are well-orders of order type $\alpha$ for any $\alpha\geq\omega$.
So that definable surjection onto $\omega_1$ is very much not a bijection either.
(And a tip for the future: whenever you write "the obvious surjection" or "the obvious bijection", you should at least write it in full details for yourself to make sure it's really obvious. Because obviously, it's not that obvious.)