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?
Best Answer
The argument makes sense in general topological spaces where singletons are closed ($T_1$ spaces I think).
If some singleton $\{x\}$ is not closed in the space $X$, then $X$ is open in $X$ while $X \backslash \{x\}$ is not.