$\textbf{First question:}$ Let $L=\{R\}$ be a language consisting of one unary relation symbol. Show that there are exactly $\aleph_0$-many countable $L$-structures up to isomorphism.
My (attempted) solution is as follows: Let $\mathcal{M}=(M,R^{\mathcal{M}})$ be an $L$-structure. Since $R$ is a unary relation symbol, the basic relation $R^{\mathcal{M}}$ is also unary. But a unary relation is just a subset of $M$. Any countable $L$-structure is isomorphic to $\omega$ or some $n \in \omega$. So the set of all countable $L$-structures up to isomorphism is $$\{(n,E):n \in \omega,E \subseteq n\} \cup \{(\omega,F):F \subseteq \omega\}$$
We first look at the set one the left hand side. It is of cardinality $\aleph_0$. If I can show the set on the right hand side is also of cardinality $\aleph_0$ then we are done. But how?
$|\mathcal{P}(\omega)|=\aleph_1$ hence the cardinality of the set on the right is $\aleph_1$. $\textbf{What did I do wrong?}$
$\textbf{Second question:}$ Let $L=\{R\}$ be a language consisting of one unary relation symbol. How many $L$-structure of size $\aleph_1$ are there?
I was trying to use the similar argument as in my first question, but in the first question, we consider countable structures and we know every countable set is isomorphic to $\omega$ or some $n \in \omega$. But if we consider structures of cardinality $\aleph_1$, I don't know the particular set they are isomorphic to, so I can't use the argument as before. What should I do?