Proof that the set of all prime numbers is not open I'm having some trouble understating this proof. The objective of the proof is to prove that the set $\mathcal{P}$ of all prime numbers is not open in the euclidean topology:

We shall now prove that $\mathcal{P}$ is not an open subset of $\mathbb{R}$. The proof goes by contradiction. Assume that it is an open subset, then given any $p_k \in \mathcal{P}$, there exists and open subset $(a,b) \subset \mathcal{P}$ such that $p_k \in (a,b)$. Let $\epsilon = \min(\{p_k-a,b-p_k\})$. By Archimedean property, we have that there exists a positive integer $n \geq 2$ such that $\frac1n \lt \epsilon$. Now we have $p_k \pm \frac1n \in (a,b)$ but $p_k \pm \frac1n \notin \mathcal{P}$. CONTRADICTION. Hence, $\mathcal{P}$ is not an open subset of $\mathbb{R}$.

My main question is about the Archimedean property. Why do they use it here? They state that because of the Archimedean property, $\exists n \geq2:\frac{1}{n}<\epsilon$, but because $\lim \frac{1}{n}=0$ (meaning that we can get arbitrarily close to $0$ as we want), isn't that obvious? Is it necessary to use the Archimedean property? And what's the purpose of that $\epsilon$ that they defined?
 A: If you want to prove that $\lim_{n\to\infty}\frac1n=0$, the natural way of doing it is to use the Archimedian property. Therefore, there is no big difference between those two options. And the numer $\min\{p_k-a,b-a_k\}$ is the distance from $p_k$ to $a$ if $a$ is closer to $p_k$ than $b$ and it is equal to the distance from $p_k$ to $b$ if $b$ is closer to $p_k$ than $a$. So, with this choice of $\varepsilon$, if $|x-p_k|<\varepsilon$, then $x\in(a,b)$ too.
A: You ask about the purpose of $\epsilon$. It's purpose is that it gives you a "cushion" for adding something to $p_{k}$ and staying in the interval $(a,b)$. In other words, if I take any number $0<\delta<\epsilon$, then $p_{k}+\delta\in (a,b)$ by definition of $\epsilon$.
Now the road to a contradiction is to choose a $\delta>0$ satisfying two things:

*

*The first is that you want $\delta<\epsilon$. Like we already discussed, you need this to ensure $p_{k}+\delta\in (a,b)$ so that you know $p_k+\delta\in\mathcal{P}$.


*The second thing you want is $\delta<1$. Why? Because then $p_k+\delta$ will not be an integer and so it can't be in $\mathcal{P}$ (a subset of the integers).
These two things will contradict each other so the main question now is: How do I know there is a real number $\delta>0$ that is both less than $\epsilon$ and less than $1$?
The proof you quote uses the the archimedean property to find such a $\delta$. The archimedian property tells me there is an integer $n$ that is bigger than $1$ and bigger than $1/\epsilon$. So if $\delta=\frac{1}{n}$ then $\delta<1$ and $\delta<\epsilon$. You say that this seems obvious and maybe you're right. But at some point or other one has to explain why its obvious that $\frac{1}{n}\to 0$ using the way integers are defined.
You're right that the proof comes off a little overdone. For one thing you only need to find one number in $(a,b)$ that is not in $\mathcal{P}$ to get a contradiction. The proof you quote finds two: $p_{k}+\frac{1}{n}$ and $p_{k}-\frac{1}{n}$. In other words, if $\epsilon$ is defined as in the proof and $0<\delta<\epsilon$ then you actually know $p_{k}+\delta\in (a,b)$ and $p_{k}-\delta\in(a,b)$. But this is more than sufficient and in reality you would only need to take $\epsilon=b-{p_k}$ to do run the argument I just described.
