Epsilon Neighborhoods of the Rationals

Question

What is meant when someone discusses an epsilon neighborhood of $\mathbb{Q}$?. Naturally the rationals are dense in $[0,1]$, so what is the epsilon neighborhood? More importantly, what does the following series add up to, $$\sum_{\substack{ 0 \leq x \leq 1, \\ x\in\mathbb{Q} }} \epsilon (x),$$ where $\epsilon(x)$ denotes the length of the interval $( x - \frac{\epsilon}{2}, x + \frac{\epsilon}{2})$.

Answer 1

It's hard to interpret your question.

If it is interpreted as a sum then (if $\epsilon$ is only allowed to be rational and positive - this is needed for the normal method of infinite series which concerns itself with countable sets), then it diverges. This can be seen easily. Take $x=\frac{1}{2}$ and consider some list of rational epsilon with $\epsilon\in (1/2,1)$. There are a countably infinite number of these $\epsilon$ and so the sum of them all, since they're all greater than $1/2$, is also infinite. Since the series is a positive one, this shows that the whole series of all $x$ and $\epsilon$ considerable diverges.

If it's interpreted as a union of all $N_\epsilon (x)$ where $x\in [0,1]$, $x\in \mathbb{Q}$, $\epsilon \in \mathbb{R}$, and $N_\epsilon (x)\subset [0,1]$, it equals $(0,1)$.* This is because the infinite union of open sets is also open. Therefore the union is open. So if $0$ were an element of the union, you would also include elements less than $0$, but we already supposed no interval/neighborhood contains elements less than zero. (same with $1$). "The contradiction establishes the theorem." ^^

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*($N_\epsilon(x)$ is "the neighborhood radius epsilon around x". Note that it varies by a factor of 2 from your original post, but that the definitions leave the problem essentially the same. I'm just treating $\epsilon$ as a radius instead of a diameter of the neighborhood.)

(please comment to let me know if I interpreted your question correctly. The second one is probably what your teacher meant? I find it the most interesting, at least.)