A question on $\omega$ Let $S=\{A \subset \omega: |A|=\omega\}$. What is the cardinality of $S$? 
If I may ask more: Let $T=\{A \subset 2^\omega: |A|=\omega\}$. then what is the cardinality of $T$?
Thanks for any help.
 A: The set $S$ has the same cardinality as $\mathcal{P}(\omega)$ (which is the cardinality of the continuum.)  To see this, notice that $S \subset \mathcal{P}(\omega)$, and on the other hand, you can construct an injection $f: \mathcal{P}(\omega) \to S$.  Hint: put all the odd numbers, say, into $f(A)$ to make it infinite, but make sure that $A$ can be recovered from $f(A)$ by looking at the even numbers.
The cardinality of $T$ is also the continuum.  Considering singletons shows that it's at least $2^\omega$ (which is the cardinality of the continuum.)  To see that it's at most $2^\omega$, you can construct a surjection from $(2^{\omega})^\omega$ to $T$, and then use the fact that $(2^{\omega})^\omega$ has the same cardinality as $2^\omega$ because $\omega \times \omega$ has cardinality $\omega$.
By the way, proving the last statement that $|T| \le |2^\omega|$ requires something like the Axiom of Choice to guarantee that given a surjection, there is an injection going the other way.
A: You can show that the cardinality of $\{A \subseteq \omega: A \text{ is finite}\}$ is $\aleph_0$, so  $\{B \subseteq \omega: |B| = \aleph_0\}$ is $2^{\aleph_0}$. As to  $\{A \subseteq 2^\omega: |A| = \aleph_0\}$, $2^{\aleph_0} = \{A \subseteq 2^\omega: |A| = 1\} \leq |\{B \subseteq 2^\omega: |B| = \aleph_0\}| \leq (2^{\aleph_0})^{ \aleph_0} $.
