I'm working on Question #5.v in chapter 3.1 of Sidney Morris's Toplogy without Tears.
I'm supposed to prove that $\mathbb{R}$ has $2^{\mathfrak{c}}$ distinct uncountable dense subsets.
I think my method gets at least halfway there.
Here's what I've done:
Note that $|\mathbb{I}| = |\mathbb{I}^{+}| = \mathfrak{c}$
$\forall\alpha\in \mathbb{I}^{+}, $ let the set $B_\alpha = \{\mathbb{R}^{-}\cup\mathbb{Q}^{+}\cup\{\alpha\}\}$
Then, each $B_\alpha$ is distinct, uncountable, and dense in $\mathbb{R}$, and there are $\mathfrak{c}$ of them indexed to the positive irrationals.
Now, consider that $|P(\mathbb{I}^{+})| = 2^{\mathfrak{c}}$, where each element of the power set is a distinct set of positive irrationals.
Since each element of the power set can be mapped 1-1 to some union of $B_\alpha$'s, we have $2^{\mathfrak{c}}$ distinct uncountable dense subsets of $\mathbb{R}$.
If this is correct, I have two further questions:
1) My construction identifies $2^{\mathfrak{c}}$ of required type of subset. Am I done? That is, should I be trying to determine whether there are exactly $2^{\mathfrak{c}}$ or more?
2) Is there a cleaner/more elegant way to construct $2^{\mathfrak{c}}$ subsets of the required type?
$\mathbb{R}$ has $2^{\mathfrak{c}}$ subsets. So yeah, there can't be more in the sense of cardinality.
By "required type" you mean "dense"? If not then obviously that totally depends on what "required type" means and it is a case-by-case study.
If yes, then your construction can be simplified: for any subset $A\subseteq\mathbb{R}\backslash\mathbb{Q}$ the subset $\mathbb{Q}\cup A$ is dense. And $\mathbb{Q}\cup A=\mathbb{Q}\cup B$ if and only if $A=B$ (under the assumption that both $A,B\subseteq\mathbb{R}\backslash\mathbb{Q}$). Since $\mathbb{R}\backslash\mathbb{Q}$ is of cardinality $\mathfrak{c}$ then you're done.