The Topology of the Lorentz group in 1+3 dimensions

This is a question of mathematical nature. Let the Lorentz group $O(1,3)$ be defined as a matrix group.

$$\text{O}(1,3) =\{\Lambda\in M_4 (\mathbb{R})| \Lambda^T \text{diag}(+---) \Lambda = \text{diag}(+---)\}$$

One defines the supremum norm on $M_4 (\mathbb{R})$ as:

which turns $M_4 (\mathbb{R})$ into a topological space in the metric topology induced by the norm. Can this supremum norm be made particular for Lorentz matrices? I guess it would, right? It would immediately follow that the Lorentz group is a topological space in the subspace topology of $M_4 (\mathbb{R})$.

Yes, $\mathrm{O}(1,3)$ is a topological subspace of $M_4(\mathbb{R})$ under this construction.
That said, there's nothing special about the topology induced by supremum norm in this regard; any subset $S$ of a topological space $X$ inherits a natural subspace topology from its "parent" space. You could define some weird topology on $M_4(\mathbb{R}) \sim \mathbb{R}^{16}$ which is a product topology between $n$ copies of $\mathbb{R}$ with the indiscrete topology and $16-n$ copies of $\mathbb{R}$ with the discrete topology, and it would still define a topology on $\mathbb{O}(1,3)$.
[EDIT: my original answer discussed whether the supremum norm might imbue the resulting topological space with non-physical structure; but as Valter Moretti pointed out in a comment, all norm topologies on a finite-dimensional vector space such as $\mathbb{R}^{16} \sim M_4(\mathbb{R})$ are equivalent, so this point is somewhat moot.]
• "By using this norm, you may be imbuing the space of matrices with a structure that isn't physically meaningful. This may or may not be problematic depending on what you do next in your derivation. " Related to this part, once the topology is defined, I can show that the Lorentz group is a non-compact topological group, by having a further subset of it (the set of Lorentzian boosts in one direction), endowed with the subspace topology, homeomorphically mapped into $\mathbb{R}$. – DanielC Oct 25 '17 at 15:07
• All norms on $\mathbb R^n$ produces the same topology. So it happens in particular on $M(m, \mathbb R)$ and their subsets. Using either the operator norm as in OP's post or the standard Euclidean norm leads to rhe same topology... – Valter Moretti Oct 25 '17 at 18:35