Subset of $\Bbb R^2$:
My book says that non-empty finite point sets are closed. Why is this?
Since it is a finite point set, it necessarily has no limit points within it, since every neighborhood of a limit point has infinite many points in it.
So a nonempty finite point set,$E$, cannot have a limit point, and this implies that all limit points of $E$ are in $E$?
Suppose $x_n \to x$, with $x_n \in E$, where $E$ is finite. Let $L = \{e \in E | x_n = e \text{ infinitely many times}\}$.
Since the limit is unique, we see that $L$ is a singleton, that is $L=\{e\}$ for some $e$. Then we see that we must have $x_n = e$ for all $n \ge N$, where $N$ is some index. Hence $x \in E$ and so $E$ is closed.
Another way follows this:
Sets consisting of a single point are closed.
The union of a finite number of closed sets is closed.
Still another way:
Let the set be $S=\{ p_1, \dots, p_n \}$ and consider the function $f(x)=d(x,p_1)\cdots d(x,p_n)$, where $d(x,y)$ is the distance between points $x$ and $y$. Then $f$ is continuous and $S$ is its zero set, and so is closed.
Another way to show it is the following. Let $E = \{x_1, \ldots, x_n\}$ and let $p \in E^c$. Let $dist_i = d(p,x_i)$ for each $i \in \{1,\ldots, n\}$ and let $m = \min \{dist_1, \ldots, dist_n\}$. We know $m > 0$ and there is some $0< \epsilon < m$. Then the ball $B_\epsilon(p) \subseteq E^c$.
Let $\delta$ be the minimum distance between different points in the set. $\delta$ > 0 but a Cauchy sequence must eventually consist of points within distance less than $\delta$ of each other. Thus the only Cauchy sequences of a finite set are sequences that are eventually constant. So the only limits of sequences of the finite set are the points in the set itself. Therefore the set is its own closure so it is closed.
