I'm trying to show that if we have a non-empty finite subset of $\mathbb{R}^{2}$ it is not an open set but it is bounded.
I have managed to show it is closed. Also, I'm aware that singleton sets are closed.
Open:
Let $ A = \{x_{1}, x_{2}, \dots, x_{k}\}$ be a non-empty finite subset of $\mathbb{R}^{2}.$ To show it is not an open set, I am trying to show that if we draw a Ball around a point in $A$ it won't be an interior point.
Let $ r > 0$ and consider $B_{r}(x_{1}) = \{y \in A : d(y, x_{1}) < r\}.$ But I'm not too sure how to proceed. I've read that this shouldn't be able to contain any open balls, but I can't see why?
Bounded: Since this is a finite set, it will have a supremum and an infimum. Would the boundedness be as simple as the fact that every point in the finite set will be between these?
Thanks.
Best Answer
Since $A$ is finite, the number $$S = \min\{d(x_i,x_j) \mid i \ne j, x_i, x_j \in A\} $$ is positive. For any $0 < s \le S$, letting $B(x_i,s)$ be the open ball around $x_i$ of radius $s$, it follows that $A \cap B(x_i,s) = \{x_i\}$. But surely $B(x_i,s)$ contains a point other than $x_i$ (can you explicitly describe such a point?), and therefore $B(x_i,s) \not\subset A$.
Here's the contradiction: If there exists $r > 0$ such that $B(x_i,r) \subset A$ then, letting $s = \min\{S,r\}$, it would follow that $0 < s \le S$ and that $B(x_i,s) \subset A$.