Prove that if $F$ is any finite subset of $\mathbb{R}$, then $\mathbb{R}-F$ is an open set.
I know this must be true because by removing the elements in $\mathbb{R}$ that are also in $F$ we are essentially creating a finite set of unions of open intervals which would be open. My question is how exactly would I formalize such a proof?
If $F$ if finite, then it is discrete and bounded. Then $F$ must be a finite union of points in $\mathbb{R}$. Call these points $P_1, \dots , P_n$. Then $\mathbb{R} \setminus F$ is just the finite intersection of all $\mathbb{R}\setminus\{P_i\}$ for all $i$. So it suffices to simply show that the real line punctured at a single point $P_i$ is open. Taking any arbitrary point in $\mathbb{R}\setminus\{P_i\}$ there exists an open ball of positive radius around this point that does not include $P_i$. This proves our claim.
To start with, suppose $F = \left\lbrace x_1, x_2, \cdots, x_n \right\rbrace \subseteq \mathbb{R}$ be your finite set. Now, for any $x \in \mathbb{R} \setminus F$, let $r_i = \left| x - x_i \right|$. Since $x \notin F$, $r_i > 0$ for each such $i$.
Now, since we have only finitely many such $r_i$, we can take $r = \min\limits_{1 \leq i \leq n} \left\lbrace r_i \right\rbrace$ and then construct the open interval around $x$ as $\left( x - \dfrac{r}{2}, x + \dfrac{r}{2} \right)$. Clearly, none of the elements of the set $F$ can be in this open ball otherwise the element inside of this interval will have a distance less than $r_i$ which will contradict the very definition of $r_i$.
Therefore, for every element in $\mathbb{R} \setminus F$, we get an interval around this element which is completely contained in $\mathbb{R} \setminus F$. Hence, $\mathbb{R} \setminus F$ must be open.
Let $F=\{x_1,\dots,x_n\}$ and let $G_i=\mathbb{R}\setminus \{x_i\}$ Then $G_i$ is open and $$\mathbb{R}\setminus F=\bigcap_{i=}G_i$$ is a finite intersection of open sets.
$F=$ {$ x_1,x_2,....x_n$ }, $x_i \in \mathbb{R}$.
$F=\bigcup_{i=1}${ $x_i $}.
{$ x_i $}, $i=1,2,.. n$ are singletons, closed in $\mathbb{R}$ (Why?).
The finite union of closed sets is closed .
$F$ is closed, hence $\mathbb{R}$ \ $F$ is open.
