Open/closed sets, transcendental numbers... Consider the metric space ℝ with the absolute value metric, d(x,y)=|x-y|. 
I need to prove whether transcendental numbers are open, closed, or neither. 
I'm stuck on how to approach this. Since there is not much information on transcendental numbers, I thought maybe I can use the algebraic numbers. So, if $S$ is the set of all transcendental numbers, I consider $S^c$ which is the complement of $S$, i.e. the set of all algebraic numbers. Now I consider a polynomial $$p(x)=\sum_{i=0}^na_ix^i$$
Then the algebraic numbers will be {x$\in$ℝ|p(x)=0}. Is this a correct line of reasoning? Any hints in the right direction?
Thank you. 

The algebraic numbers are the set of all zeros to some polynomial equation, but knowing that doesn't really help us much.
We can show $S$ is not open by showing that some point is not an interior point. To do this, note that $\mathbb Q$ is dense in $\mathbb R$ and take open balls around the transcendental $e$.
We can show that $S$ is not closed by showing that some limit point of $S$ is not in $S$. To do this, note that each number in the sequence $(\frac{e}{n})$ is transcendental and this converges to $0$.
