Answer to Question #217105 in Differential Geometry | Topology for Prathibha Rose

Question

Let (x,d) be metric space and A proper subset of X .Define the closure of a set A .consider the usual metric space (R^n,d) .let A = {(x₁,x₂,.......x_n): x_i element of Q}

Answer 1

There is three possible definitions of a closure of a set $A$ (which are equivalent for a metric space) :

• the smallest closed set which contains $A$,
• the set of all possible limits in $X$ of sequences of elements of $A$,
• the set of points $x\in X$ such that any neighbourhood of $x$ intersects $A$.

For a space $\mathbb{R}^n$ with a usual metric $d$ and a set $A$ defined as $x\in A \leftrightarrow (x=(x_1,...,x_n) \wedge \forall i\: x_i \in \mathbb{Q})$ the closure of $A$ is the entire space $\mathbb{R}^n$. We can prove it, using the second definition. Let $x\in\mathbb{R}^n$ with $(x_1,..., x_n)$ its components. By density of $\mathbb{Q}$ in $\mathbb{R}$, there are sequences $(x_i^k)_{k\in \mathbb{N}}$ with $x^k_i \in \mathbb{Q}$ and $x_i^k\overset{k\to+\infty}{\to} x_i$ for all $1\leq i\leq n$. The sequence in $A$ defined as $((x_1^k,...,x_n^k))_{k\in\mathbb{N}}$ converges to $x=(x_1,...,x_n)$, as all coordinates of $x^k$ converge to a respective coordinate of $x$. Therefore, $x\in \bar A$ and as this point is arbitrary, we have $\bar A=\mathbb{R}^n$.