It is well know that $SO(n)$ is connected and $O(n)$ has two connected components: $O^+(n)=\{A\in O(n):\det A=+1\}$ and $O^-(n)=\{A\in O(n):\det A=-1\}$.
In what book can I find this property?
It is well know that $SO(n)$ is connected and $O(n)$ has two connected components: $O^+(n)=\{A\in O(n):\det A=+1\}$ and $O^-(n)=\{A\in O(n):\det A=-1\}$.
In what book can I find this property?
Suppose $A \in O(n)$. Then, by definition, $A^tA=\mathbb{I} $, where $A^t$ represents the transpose of the matrix $A$ and $\mathbb{I}$ represents the identity. Taking the determinant of this equation, we get
\begin{align*} \text{det}(A^t)\text{det}(A) &=1 \\ \Rightarrow (\text{det}(A))^2 &= 1 \\ \Rightarrow \text{det}(A) &= \pm 1 \tag{1}\\ \end{align*}
Therefore, we see that $$ O(n)=O^-(n)\ \coprod \ O^+(n) \tag{2}$$ where $\coprod$ stands for disjoint union.
Now we know that the determinant is a continuous map. Therefore, the pre-image of an open set in $\mathbb{R}$ must be open in $O(n)$. Consider two disjoint open intervals $B_\epsilon(-1)\equiv(-1-\epsilon,-1+\epsilon)$ and $B_\epsilon(+1)\equiv(1-\epsilon,1+\epsilon)$ for some small $\epsilon>0$.
By continuity of the determinant, $\text{det}^{-1}(B_\epsilon(-1))$ and $\text{det}^{-1}(B_\epsilon(+1))$ are open in $O(n)$. But observe that, we have precisely:
$$ \text{det}^{-1}(B_\epsilon(-1))=O^-(n) \tag{3}$$ $$ \text{det}^{-1}(B_\epsilon(+1))=O^+(n) \tag{4}$$
Therefore, (2), (3) and (4) prove that $O(n)$ is not connected and that there are at least two connected components of $O(n)$. In order to prove that $O(n)$ has exactly two connected components, one further needs to prove that $O^\pm (n)$ are connected, in fact arcwise-connected.
I would recommend Matrices: Theory and Applications by Denis Serre. Read Chapter 10 where he discusses all this in great detail. Moreover, the book is very beginner-friendly I guess. If not, it is very readable nonetheless.