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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.

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up vote 5 down vote accepted

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 $0$.

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Hint 1: Can a countable set be open? Algebraic numbers are countable...

Hint 2: $\mathbb Q \subset S^c$ and $\mathbb Q$ is dense in $\mathbb R$.

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