Complement of any finite set in metric space. I am asked to show that in a metric space the complement of any finite set is open. That means I have to show that finite set is closed. I just tried to cope up with this problem using closure definition but dont know whether my approach is correct or not.  I know that the union of $A$ and $A'$ (the set of all interior points of $A$) comes to be $A$ if I take $A'$ to be empty or $A'$ to be $A$, so that I get  (closure of $A$) $ = A$ and by using the statement (closure of $A$)$=A$ iff $A$ is closed, I get the proof done.
My question is can I apply this definition here. Sorry if I am asking something weird.
I am just asking whether or not should it is absolutely proper to use that specific definition, thats all. aso, I think my question has some additional questions about using the definition.
 A: In short, yes if something in math has a property that has a definition, then the definition can be used to prove that property. It would be a bit odd for a set to be closed, yet you would be unable to prove it is closed using the definition. That being said, there are multiple equivalent definitions of what it means to be closed and you mentioned two of them. Sometimes it is definitely easier to use one definition over another, depending on the nature of the problem and the experience of the problem-solver. I personally lean toward showing that the complement of a finite set is open for this problem. The fact that we have a metric space $X$ that allows us to measure distance in the space lends itself to the following argument:
Let $A \subset X$ be a finite set, $A = \{a_1,a_2,\ldots a_n\}$ and let $x \in X\setminus A$. Since $x$ is not one of the $a_i$'s it must be a distance of $d(a_i,x) = \varepsilon_i>0$ away. In this way we can build another set $D=\{\varepsilon_i/2\}_{i=1}^n$, that is all the halved-distances from $x$ and each $a_i$. Since $D$ is finite we can define $\min\{D\} = \varepsilon.$ Now the open ball $B_\varepsilon(x) = \{y \in X: d(x,y) < \varepsilon \}$ is an open set that contains $x$ and has been constructed precisely so that $B_\varepsilon(x) \subset X\setminus A$. Since $x$ was arbitrary, $X\setminus A$ must be open.
A: It's perfectly fine to show that a finite set is closed because it's equal to its own closure. But to compute the closure, you have to take the set of all boundary points of $A$ union $A$, not the set of interior points! So you need to show that a finite set has no boundary points, in other words, that no sequence contained in a finite set converges outside the set. That's where the work is here. As a hint: try to characterize the sequences in a finite set which converge.
A: Hint: All metric spaces are Hausdorff. 
