TITLE: Does removing finitely many points from an open set yield an open set?
QUESTION [17 upvotes]: Removing finitely many point from an open set in $\mathbb{R}^n$ gives an open set.  Is this true in general for any space?
My intuition is that this is the case, however, how does one (dis)prove this?
The only idea that comes in mind is that, since singletons are closed and unions of closed sets are closed, a union of singletons is closed.
Now, let $U$ be an open set.  Then the complement of $U$ is closed, and any point $x$ removed from $U$ is a singleton that is "unioned" with the closed complement of $U$ to form a bigger closed set $C$.  $C$ is the complement of $U\setminus\{x\},$ so $U\setminus\{x\}$ is open.  
One can thus repeat this argument inductively for any finite number of removed points. 
Does this make sense?

REPLY [14 votes]: The reason this works in $\Bbb R^n$ is that a set consisting of a single point is always closed.  Then you can understand $$G\setminus\{ x_1, x_2, \ldots, x_k \}$$ as $$G\cap (\Bbb R^n \setminus \{x_1\}) \cap (\Bbb R^n \setminus \{x_2\})\cap\ldots \cap (\Bbb R^n \setminus \{x_k\}).$$ This is a finite intersection of open sets, so is open.
In general spaces, a singleton set $\{x_1\}$ may not always be closed.  A space in  which every singleton is closed is called a $T_1$ space.  All metric spaces are $T_1$, but it is easy to construct quite simple spaces that are not $T_1$.  For example, let $S=\{a,b,c\}$ and let the open sets be $S, \{a,b\}, \{a,c\}, \{a\}, \varnothing$. Then $\{c\}$ is closed but $\{a\}$ is not, and indeed if you remove the single point $a$ from the open set $\{a,b\}$ you get $\{b\}$, which is not an open set.