Removing points from an open set Let $A$ be an open set in $\mathbb{R}^k$. Show that if we remove a finite number of points from $A$, the resulting set is still open. Give an example where we remove an infinite sequence of points $(a_1, a_2, \dots)$ and the resulting set is no longer open.
 A: Let $A \subseteq \Bbb R^p$ be open. That is given any $x \in A$ there is a neighbourhood of $x$ entirely contained in $A$. We can interpret this to say that given any $x$ there is an open ball $B_x(r)$ centred at $x$ with radius $r \gt 0$ such that $B_x(r) \subseteq A$. Suppose the removed finite number of elements are in $\{x_1, x_2, ... x_n\}$. Pick $R_x = \text{Min}\{ r, \left|{\left|{x - x_1}\right|}\right|, \left|{\left|{x - x_2}\right|}\right|, .., \left|{\left|{x - x_n}\right|}\right|\}$, then the neighbourhood,  $B_x(R_x)$ of $x$ is contained in $A$. 
Let $y \in A \setminus \{x_1, x_2, ... x_n\}$. Then $R_y$ exists and $B_y(R_y) \subseteq A \implies A \setminus \{x_1, x_2, ... x_n\}$ is open in $\Bbb R^p$
As for the example $(0, 1)$ is open in $\Bbb R$. But $(0,1) \setminus \Bbb Q$ is not open as long as you are  willing to give that $\Bbb Q$ is countable and hence $\Bbb Q \cap (0, 1)$ can be represented by an enumerated sequence. 
A: I know this is quite an old post but I just wanted to add another possible way to see this. 
In $\mathbb{R^k}$, any singleton set is closed (using the definition that a set is closed if it contains all its limit points, this is vacuously true). Thus the complement of a singleton set is open by definition.
Given an open set $A \subset \mathbb{R^k}$, say we remove the points $\{x_1, \dots x_n\}$ from $A$.
Now the resulting set can be written as, 
$$A\setminus \{x_1,\dots x_n\} = A \cap\{x_1\}^c \cap \dots \cap \{x_n\}^c$$
This is a finite intersection of open sets, thus $A\setminus \{x_1,\dots x_n\}$ must be open.
