Proof involving a vacuously true statement Let $S$ be a finite subset of a metric space. Show that it is closed.
I know a set is closed if and only if it contains all of its accumulation points. Let $x$ be an accumulation point of $S$. I want to show that $x \in S$. Since $S$ is a finite set, I know that it cannot have a accumulation point. So does this imply that $x \in S$ and I am done with the proof?
If I have made an error, could someone explain? 
 A: To prove that $S$ is closed in a metric space $M$, it is enough to show that $S$ contains all its accumulation points.
Since $S$ is finite, we can show that it has no accumulation points. Proof by contradiction: suppose that $S$ has an accumulation point $x$. Consider $r = \min |x-y|$ for all $y \in S$. Then the ball around $x$ with radius $r/2$ will contain no points of $S$. So $S$ contains no accumulation points. 
Since $S$ contains all of its accumulation points (since $S$ has none), that means $S$ is closed in $M$.
A: Suppose that finite subsets contain interior points .But for a finite subset  we can find a least distance i.e we can find a ball around any point which contains no points of the set ,so no point is an interior point.
A: This question was asked almost a year ago; I wanted, however, to propose an alternate solution.
Let $(X,d)$ be a metric space and let $S$ be a finite subset of $X$, say $S=\{x_{1}, \ldots, x_{n}\}$. A set is closed if its complement is open. So, we shall show that $X\backslash S$ is open. Pick an arbitrary point $x \in X \backslash S$. Let $r=\min\{d(x_{i},x): i=1, \ldots, n\}$. ( Note that $d(x_{i},x) \neq 0$ for each $i=1, \ldots, n$. Hence, $r > 0$. ) Then $B(x;r)$ is an open ball containing $x$ that lies entirely in $X\backslash S$. That is, $X\backslash S$ is a neighborhood of each of its points (because $x$ was arbitrary). We conclude $X\backslash S$ is open; it follows that $S$ is closed.
