# Finding closure of a set in Euclidean Space

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?

• What defintion of a closed set can you use?
– mag
Commented Oct 15, 2021 at 14:51
• A is not closed, can u elaborate? thanks Commented Oct 15, 2021 at 14:53
• Notice that if $x$ is in $A$, clearly any neighborhood of $A$ contains a point in $A$. Now what happens if you're on the boundary? Take a point $(x,y)$ such that $x^2+y^2=1$. What does a neighborhood of $(x,y)$ look like? Does it contain a point in $A$? Commented Oct 15, 2021 at 14:54
• @Saegusa , well first you have to take another thing into account, if x is not on the boundary and is not in A, Secondly i know how it looks like, and well aware of the geometric interpretation. but i can't seem to right an exact proof. Commented Oct 15, 2021 at 14:57
• You might want to use the density of $\mathbb{Q}$ in $\mathbb{R}$ and the sequential characterization of closure for subsets of the Euclidean space $\mathbb{R}^2$. Commented Oct 15, 2021 at 15:01

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 $$\mathbb{R}^2$$.