Find the cardinality of set $A$: $$A = \{x \subseteq \mathbb R : |x| < \aleph_o \}$$
I have thought about it for a while and I've come to the conclusion that the cardinality of this set will be at least continuum, because every real number singleton is in this set. However, the only upper-bound of this cardinaltiy that I can think of is $\mathbb R^{\mathbb R}$ whose cardinality is more than continuum.
Any suggestions and hints would be most appreciated.
On
Recall that the cardinality of $\mathbb{R}$ is $2^{\aleph_0}$.
We can in fact show that the set of all countable subsets of $\mathbb{R}$ is $\mathbb{R}$-sized, and then the result follows because you're just asking for a subset of those.
Note that $(2^{\aleph_0})^{\aleph_0}$ is equal to $2^{\aleph_0 \times \aleph_0} = 2^{\aleph_0}$, so the number of countable subsets of $\mathbb{R}$ is just the cardinality of $\mathbb{R}$ again.
This is the number of finite subsets of $\Bbb R$. For a given natural number $n$, the number of $n$-element subsets of $\Bbb R$ is bounded above by $|\Bbb R^n|=|\Bbb R|$. So $|A|\le\sum_{n=1}^\infty|\Bbb R|=\aleph_0|\Bbb R|=|\Bbb R|$.