Why is $\mathbb R$ \ ${0}$ open in R

Question

I needed to prove that $GL(n,\mathbb R)$ is a manifold. I first tried to use the determinant as a function $det:M_n \rightarrow \mathbb R $ where $M_n$ is the set of all $n\times n$ matrices in order to define open sets on $M_n$ but after that I got stuck. When I looked it up they used the determinant to define open sets but then go on to say because $GL(n,\mathbb R)=det^{-1} (\mathbb R $\ $ {0})$ and $\mathbb R $\ $ { 0}$ is open in $\mathbb R$ $GL(n,\mathbb{R})$ is open in the set of matrices and is therefore a manifold. Why is $\mathbb R $\ $ { 0}$ open in $\mathbb R$?

Answer 1

The complement of $\mathbb{R}\setminus\{0\}$ is $\{0\}$, which is closed in the Euclidian topology. Hence, it follows that $\mathbb{R}\setminus\{0\}$ is open.

Answer 2

Usually the topology on $\mathbb{R}$ is defined by: subset $U\subseteq\mathbb{R}$ is open if for any $x\in U$ there is $r\in\mathbb{R}_+$ such that the open ball $B(x,r)\subseteq U$.

With that take $U=\mathbb{R}\backslash\{0\}$ and any $x\in U$. Then $B(x,|x|)\subseteq U$. Proving that $U$ is indeed open.

Answer 3

$$\mathbb R\backslash \{0\}=(-\infty ,0)\cup(0,\infty ),$$ that is the union of two open.