Can a function $f:[0,2\pi] \rightarrow 1$ have a PDF and CDF? Consider an interval $X = [0,2\pi]$ Assume $X$ has a uniform distribution. 
Let $f: X \rightarrow 1$ be a transformation of the $X$ to a random variable $Y$ whose only member is the number $1$
Can I define a CDF for $f(X)$? 

I am having trouble answering this but I have some thoughts. I am a bit confused because it seems very weird to map all of $X$ onto a single number. Doesn't that effectively mean $\Bbb P(y = 1) = 1$ if $x\in [0,2\pi]$ but $\Bbb P(y=1) = 0 $ if $x\notin [0,2\pi]$? There aren't any other possible values for $y$ so it seems like this must be the answer. However, usually my CDF are all written such that $x$ is no longer relevant and everything is in terms of $y$, so I feel like this reasoning can't be right even though it makes sense. 

What you mean is $X$ is uniformly distributed on $[0,2\pi]$ and $Y=f(X)$, where $f(x) \equiv 1$ for all $x$. (In particular, this means $X$ is a random variable, not an interval.)
The problem with this definition is that $Y$ is not a random variable, because its pdf is exactly $$f_Y(y) = \begin{cases} 1, & y \in [0,2\pi], \\ 0, & \text{otherwise}, \end{cases}$$
which means
$$
\int_\mathbb{R} f_Y(y) \, dy = \int_0^{2\pi} 1 \, dy = 2\pi \ne 1 \cdots
$$
Can you finish this?
