# Prove that the following set $A = \{x \in \mathbb{R} \setminus\mathbb{Q}\mid x^2 \leq \frac{1}{4} \} \subset \mathbb{R}$ is complete.

A subset $$S$$ of $$\mathbb{R}$$ is complete if every Cauchy sequence consisting of elements of $$S$$ converges to an element of $$S$$.

Prove that the following set $$A = \{x \in \mathbb{R} \setminus\mathbb{Q}\mid x^2 \leq \frac{1}{4} \} \subset \mathbb{R}$$ is complete.

Note: I don't want the solution (at least not eight now) but I want to run past people my idea of how to prove this.

IDEA:

I feel that this set is not complete. I'm basing the idea off of the same notion of how the rationals $$\mathbb{Q}$$ are not complete. Particularly I could have a sequence converging to $$\sqrt{2}$$. In this problem I want to use that same approach. Based on the conditions of my set

$$\frac{-1}{2} \leq x \leq \frac{1}{2}$$

So I was thinking that I could create a sequence that would converge to $$\frac{1}{2}$$. And by virtue of it converging it means it is Cauchy. To create this sequence I would use the idea that the product of an irrationl and rational number is an irrational number. An idea of what I would like the sequnce terms to look like would be something of this vein:

$$x_n = \Bigg(\frac{n-1}{n}\Bigg)\Bigg(\frac{1}{\sqrt{2}}\Bigg)$$

I know this specific term doesn't work. What I was hoping to accomplish was since $$\lim_{n \rightarrow \infty}\frac{n-1}{n} \rightarrow 1$$, I could somehow use that product of terms to converge to $$\frac{1}{2}$$ What would be ideal was if I could somehow manufacture the irrational terms converge to 1 and have $$\frac{1}{2}$$ as a constant, but I don't know how I could possibly create a sequence of irrationals without having an explicit irrational value in it.

Am I on the right path? What could I do to help my situation?

If you want an explicit sequence, you could take the sequence $$a_n = 1/2 - 1/\sqrt{p_n}$$ with $$p_n$$ the $$n$$'th prime number.