Open set on $E^n$ (the $n$-dimensional euclidean space) Let $A$ be a countable set of $E^n$ (the $n$-dimensional euclidean space). Show that $A$ is not an open set of $E^n$.

Definition of open set
Let $(X, \mathcal T)$ be a topological space. A subset $U \subset X$ is called an open set  of $X$ if $U \in \mathcal T$. For $x \in X$, if $U$ is an open set and if $x \in U$, then we call $U$ a neighborhood of $x$

How to show that $A$ is not an open set of $E^n$ ?

For $n>1$ and non-empty $A$: Let $x=(x_1,...,x_n)\in A.$ If $A$ is open then for some $r>0$ the open ball $B(x,r)$ of radius $r$, centered at $x$, is a subset of $A.$ The real interval $(-r+x_1,r+x_1)$ is uncountable and  $$A\supset B(x,r)\supset (-r+x_1,r+x_1)\times (x_2,...,x_n)$$ which is  uncountable.
