Finding closure of a set in Euclidean Space

Question

In Euclidean Space $R^2$, A : {(x,y) ∈ $Q^2 : x^2 + y^2 < 1$ }. I'm trying to prove that the closure of A is $\overline A$ : {(x,y) ∈ $R^2 : x^2 + y^2 \le 1$ }. In other words I have to prove that any nbd of any point in $\overline A$ contains a point in A. Any help?

Answer 1

For $x \in \mathbb{R}^2, r > 0$, set $B(x, r) = \{y \in \mathbb{R}^2 : \lVert y - x \rVert < r\}$, the open ball of radius $r$ centered at $x$.

Using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ to prove that $\mathbb{Q}^2$ is dense in $\mathbb{R}^2$, i.e. $\overline{\mathbb{Q}^2} = \mathbb{R}^2$. Also prove that $\overline{B(0, 1)} = \{x \in \mathbb{R}^2 : \lVert x \rVert \leq 1\}$ (this is true in any normed vector space, not just $\mathbb{R}^2$).

You want to show that $\overline{B(0, 1) \cap \mathbb{Q}^2} = \overline{B(0, 1)}$. It suffices to show that $\overline{B(0, 1) \cap \mathbb{Q}^2} \supset B(0, 1)$. But this is easy to show using the density of $\mathbb{Q}^2$ in $\mathbb{R}^2$.