What is the topology on $O(3,1)$? I've read that there are four connected components of $O(3,1)$. Then, if I'm not mistaken, if $A \in O(3,1)$, then to determine which connected component $A$ is in we look at whether $\text{det}(A)= \pm 1$ and look at the sign of the first entry of $A$ is either greater than or equal or less than or equal to 1. My question is what topology is given to $O(3,1)$ such that these are the components?

The real $n \times n$ matrices $M_n(\mathbb{R})$ form a normed vector space. (Take the operator norm or view a matrix as a vector in $\mathbb{R}^{n^2}$ and take the usual norm there.) For finite-dimensional spaces all norms are equivalent and so they induce the same topology. Thus an open set in $M_n(\mathbb{R})$ is a union of open balls with respect to the metric $d(A,B) = \|A-B\|$, where $\| \cdot \|$ is any norm. The topology on $O(p,q) \subset M_n(\mathbb{R})$ ($p+q = n$) is just the subspace topology.