Question on the existence of finite open subsets in $\mathbb{R}^{k}$ Dear reader of this post,
can any subset $S \subset \mathbb{R}^{k}$, with the euclidean distance for $x,y \in \mathbb{R}^{k}$, be open and contain only finitely many elements? 
I am looking forward for your reply! 
 A: Let's assume that $S$ is not empty (As Daniel Fischer pointed out in the comments, this is trivially open and finite).
If $S$ is finite, then for every $s\in S$, there exists a $\delta>0$ such that $d(s,s')>\delta$ for all $s'\in S\setminus\{s\}$ (we can set $\delta=\frac{1}{2}\min\{d(s,s')|s'\in S\setminus\{s\}\}$). So every point in $S$ is isolated. It follows that $S$ can not be open as every open ball around $s$ contains points not in $S$.
A much easier way to show this would be to just note that the cardinality of any open ball is infinite and so can not be a subset of a finite subset of $\mathbb{R}^k$.
A: Another approach is via connectedness: Assume $F$ is a finite subset of $\Bbb R^k$. Then it is a union of singletons $\{x_1\},...,\{x_n\}$. Since each $\{x_i\}$ is closed, so is the finite union $F$. Now, $\Bbb R$ is known to be connected, and a product of connected sets is connected. Therefore, $F$ cannot be open and closed at the same time.
More generally, this shows that in a connected $T_1$-space open sets cannot be finite unless they are empty.
