Show that a finite subset of a metric space has no limit points and is therefore a closed set.
A subset of a metric space is closed if and only if it contains its limit points.
To be a limit point any neighborhood needs to contains infinitely many points within the set.
Thus, a finite subset cannot have any limit points.
Thus, the set of limit points is the empty set.
But, the finite subset contains the empty set.
Thus, the finite subset is closed.
This is an alternate solution to TPHs...it is a little more formal. Note, you may want to wait for confirmation of my solution by a more knowledgable member
Let be a metric space and let . Now the only possibilities for a limit point of is the point , but note that since there is only one point no matter how small you make an open ball around there will not be another point of , thus a singleton in has no limit points, thus it is closed and consequently its complement is open. So now let another set where is finite. It is clear that we will now use the following lemma
Lemma: The intersection of any finite number of open sets is itself open.
Proof: Let be open, then let . This implies that , and since each is open there exists an open ball corresponding to each such that . So now let , then
Using the above lemma and the fact that the complement of any singleton in is open we can see that the complement of any finite point set is open, thus the finite set is closed.