If we take a finite number of points away from Q is it still dense in R usual topology Is Q less a finite number of points dense?
My attempt:  any intersection of any open non-empty set U with Q is infinite since U has infinite many rationals.   Therefore, if we remove at most a finite amount of points from Q then the intersection with any non empty set U will still be non-empty (ie infinite number of points minus at most finitely number of points still leaves infinite amount of points (but in particular at least one point therefore non empty).
Is this correct?
 A: If $X$ is a $T_1$ space (so finite subsets are closed in $X$, this is true in all metric spaces, so certainly in $\Bbb R$..) without isolated points (so no $\{x\}$ is an open set in $X$, also true in $\Bbb R$..) then for any dense set $D$ of $X$ and any finite set $F \subseteq D$, $D\setminus F$ is still dense in $X$.
Proof: Let $O$ be open and non-empty in $X$. Then $O\setminus F=O \cap F^\complement$ is also open (as $X$ is $T_1$ and we have an intersection of two open sets) and $O\setminus F$ is non-empty because $X$ has no isolated points ($O \setminus F = \emptyset \to O \subseteq F$ and $F$ would have an interior point and thus must be an isolated point in a $T_1$ space, contradiction). So as $D$ is dense
$$\emptyset \neq (O \setminus F) \cap D = O \cap (D \setminus F)$$
showing that indeed $D \setminus F$ is dense, as $O$ was arbitrary.
So this fact holds in many spaces. Isolated points of $X$ are special here because they have to be in any dense set (so can never be removed from a dense set).
