The timelike set has exactly two convex components, described in orthonormal coordinates

For $v={\sum }_a{v}^a{e}_a$ and $w={\sum }_b{w}^b{e}_b$, bilinearity of $\eta$ together with $\eta (e_a,e_b)=\mathrm{diag}(-1,1,1,1{)}_{ab}$ from [L1] give $\eta (v,w)=-v^0{w}^0+v^1{w}^1+v^2{w}^2+v^3{w}^3$, hence $\eta (v,v)=-(v^0{)}^2+\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mid }^2$, where $\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mid }^2=(v^1{)}^2+(v^2{)}^2+(v^3{)}^2=\parallel v_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2^2$ by [L4]. The assignment $c(v):=(v^0,v^1,v^2,v^3)$ is a bijection $V\to {\mathbb{R}}^4$ by [L2]. By [L3] the timelike set is $T=\{v:\eta (v,v)<0\}=\{v:(v^0{)}^2>\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mid }^2\}=\{v:\mid v^0\mid >\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \}=T_{+}\cup T_{-}$, because $\mid v^0\mid >\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ holds exactly when $v^0>\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ or $v^0<-\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$.

Define $d(u,w):=\parallel c(u)-c(w){\parallel }_2$. By [L5] applied to the Euclidean inner product of [L4], the Euclidean norm is nonnegative, vanishes only at $0$, is absolutely homogeneous, and satisfies the triangle inequality; with $c$ a bijection [L2] this shows that $d$ is a metric on $V$ in the sense of [L6]: $d(u,w)=0$ if and only if $c(u)=c(w)$, if and only if $u=w$; $d(u,w)=d(w,u)$ by $\parallel -x\parallel =\parallel x\parallel$; and $d(u,w)\le d(u,z)+d(z,w)$ by the triangle inequality applied to $(c(u)-c(z))+(c(z)-c(w))$.

The coordinate image $c(T_{+})=\{x\in {\mathbb{R}}^4:x_0>\parallel x_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2\}$ is convex in the sense of [L7]: if $x,y\in c(T_{+})$ and $t\in [0,1]$, then the first coordinate of $z:=(1-t)x+ty$ is $z_0=(1-t)x_0+t{y}_0>(1-t)\parallel x_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2+t\parallel y_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2\ge \parallel (1-t)x_{\mathrm{v}\mathrm{e}\mathrm{c}}+t{y}_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2=\parallel z_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2$, where the strict inequality uses $x_0>\parallel x_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2$, $y_0>\parallel y_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2$ and $t,1-t\ge 0$ and the second inequality is the triangle inequality with homogeneity from [L5]; hence $z\in c(T_{+})$. The same computation with reversed inequalities shows that $c(T_{-})=\{x:x_0<-\parallel x_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2\}$ is convex.

The coordinate vector of $e_0$ is $(1,0,0,0)$ and that of $-e_0$ is $(-1,0,0,0)$, by uniqueness of coordinates [L2]; hence $e_0\in T_{+}$ and $-e_0\in T_{-}$, so both sets are nonempty. If $v\in T_{+}\cap T_{-}$, then $v^0>\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \ge 0$ and $v^0<-\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \le 0$ simultaneously, which is impossible; hence $T_{+}$ and $T_{-}$ are disjoint and $T=T_{+}\bigsqcup T_{-}$.

Let $f(v):=v^0-\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$. For all $u,w\in V$ one has $\mid u^0-w^0\mid \le d(u,w)$, because $(u^0-w^0{)}^2$ is one term of the sum $\parallel c(u)-c(w){\parallel }_2^2$, and $\mid \mid u_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid -\mid w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \mid \le \parallel u_{\mathrm{v}\mathrm{e}\mathrm{c}}-w_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2\le d(u,w)$, the first inequality from $\mid u_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \le \parallel u_{\mathrm{v}\mathrm{e}\mathrm{c}}-w_{\mathrm{v}\mathrm{e}\mathrm{c}}{\parallel }_2+\mid w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ and its symmetric partner by [L5] and the second because the last three coordinate squares form a sub-sum; hence $\mid f(u)-f(w)\mid \le 2d(u,w)$. If $f(v)>0$ and $d(u,v)<f(v)\mathrm{/}2$, then $f(u)\ge f(v)-2d(u,v)>0$; therefore $T_{+}=\{v:f(v)>0\}$ is open in the metric topology of [L9], since the ball of radius $f(v)\mathrm{/}2$ about each of its points lies in it, and the same argument with $g(v):=-v^0-\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ shows that $T_{-}=\{v:g(v)>0\}$ is open.

By step 2.1 the sets $c(T_{+})$ and $c(T_{-})$ are convex subsets of ${\mathbb{R}}^4$, hence path-connected by [L8]. Let $p:[0,1]\to c(T_{+})$ be a path in $c(T_{+})$ and put $\gamma (t):=c^{-1}(p(t))$, a map into $T_{+}$ because $c$ is a bijection; then $d(\gamma (s),\gamma (t))=\parallel p(s)-p(t){\parallel }_2$ for all $s,t$, so for every $\eta >0$ the $\delta >0$ supplied by continuity of $p$ at $t$ gives $d(\gamma (s),\gamma (t))<\eta$ whenever $\mid s-t\mid <\delta$: thus $\gamma$ is a path in the metric space $(V,d)$, and $T_{+}$ is path-connected. The same argument gives that $T_{-}$ is path-connected. By [L13] the path-connected sets $T_{+}$ and $T_{-}$ are connected subsets of $T$, and by step 2.2 they are nonempty with $T=T_{+}\bigsqcup T_{-}$.

Suppose $C\subseteq T$ is a connected subset meeting both $T_{+}$ and $T_{-}$. The sets $C\cap T_{+}$ and $C\cap T_{-}$ are open in the subspace topology on $C$ by step 2.3 and [L11], they are nonempty because $C$ meets both, they are disjoint, and their union is $C$ because $C\subseteq T=T_{+}\bigsqcup T_{-}$ by step 1.1; thus $(C\cap T_{+},C\cap T_{-})$ is a separation of $C$ in the sense of [L10], contradicting connectedness of $C$. Hence every connected subset of $T$ is contained in $T_{+}$ or in $T_{-}$.

The identity $\mathrm{cl}(T_{+})=\{v:f(v)\ge 0\}$ holds for the closure in the metric topology of [L9]: if $f(v)<0$, then every $u$ with $d(u,v)<-f(v)\mathrm{/}2$ has $f(u)<0$ by the Lipschitz bound of step 2.3, so a ball about $v$ misses $T_{+}=\{f>0\}$ and $v\mathrm{\notin }\mathrm{cl}(T_{+})$; conversely, if $f(v)\ge 0$, then for every real $t>0$ the point $v+t{e}_0$ has coordinates $(v^0+t,v^1,v^2,v^3)$ by [L2], hence $f(v+t{e}_0)=f(v)+t>0$, so $v+t{e}_0\in T_{+}$ and $d(v+t{e}_0,v)=t\to 0$. The same argument with $g(v)=-v^0-\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ gives $\mathrm{cl}(T_{-})=\{v:g(v)\ge 0\}$; in coordinates these are the displayed sets $\{v:v^0\ge \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \}$ and $\{v:v^0\le -\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \}$.

Let $\gamma :[0,1]\to T$ be a path with $\gamma (0)\in T_{+}$ and $\gamma (1)\in T_{-}$, and put $h(t):=(\gamma (t){)}^0$. Using the bound $\mid u^0-w^0\mid \le d(u,w)$ of step 2.3, for every $\eta >0$ the $\delta >0$ supplied by continuity of $\gamma$ at $t$ gives $\mid h(s)-h(t)\mid \le d(\gamma (s),\gamma (t))<\eta$ whenever $\mid s-t\mid <\delta$; hence $h$ is continuous on $[0,1]$ in the sense required by [L14], with $h(0)>0$ because $\gamma (0)\in T_{+}$ and $h(1)<0$ because $\gamma (1)\in T_{-}$. By [L14] there is $t\in [0,1]$ with $h(t)=0$; but $\gamma (t)\in T$ gives $(h(t){)}^2=(\gamma (t{)}^0{)}^2>\mid \gamma (t{)}_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mid }^2\ge 0$ by step 1.1, so $h(t)\mathrm{\ne }0$, a contradiction. Hence no path in $T$ joins $T_{+}$ to $T_{-}$.

By step 3.1 the sets $T_{+}$ and $T_{-}$ are connected, and by step 2.2 they are nonempty and disjoint with $T=T_{+}\bigsqcup T_{-}$. Let $x\in T_{+}$. By [L12] the component $C_T(x)$ is the largest connected subset of $T$ containing $x$; since $T_{+}$ is a connected subset of $T$ containing $x$, one has $T_{+}\subseteq C_T(x)$, while $C_T(x)$ is connected and contains $x\in T_{+}$, so step 3.2 gives $C_T(x)\subseteq T_{+}$; hence $C_T(x)=T_{+}$. The same argument with $x\in T_{-}$ gives $C_T(x)=T_{-}$. Therefore $T_{+}$ and $T_{-}$ are exactly the connected components of $T$, and since $T=T_{+}\bigsqcup T_{-}$ these are all of them.

If $v\in \mathrm{cl}(T_{+})\cap \mathrm{cl}(T_{-})$, then step 3.3 gives $v^0\ge \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ and $v^0\le -\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, hence $0\le 2\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \le 0$, so $\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid =0$ and $v^0=0$; all coordinates of $v$ vanish, and uniqueness of coordinates [L2] gives $v=0$. Conversely $0$ lies in both displayed sets of step 3.3, hence in both closures. Therefore $\mathrm{cl}(T_{+})\cap \mathrm{cl}(T_{-})=\{0\}$.

The descriptions of step 3.3 also show that $\mathrm{cl}(T_{+})$ and $\mathrm{cl}(T_{-})$ are convex cones and that the nonzero causal vectors are exactly the nonzero points of their union. For the cone property, if $v,w\in \mathrm{cl}(T_{+})$ and $\lambda \ge 0$, then $(v+w{)}^0=v^0+w^0\ge \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid +\mid w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid \ge \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}+w_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ and $(\lambda v{)}^0=\lambda v^0\ge \lambda \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid =\mid \lambda v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ by [L5], so $v+w,\lambda v\in \mathrm{cl}(T_{+})$, and reversing the inequalities gives the same for $\mathrm{cl}(T_{-})$. For the classification, every timelike vector lies in $T=T_{+}\cup T_{-}\subseteq \mathrm{cl}(T_{+})\cup \mathrm{cl}(T_{-})$ by step 1.1, and every null vector $v$ has $\mid v^0\mid =\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$ by step 1.1, hence lies in $\mathrm{cl}(T_{+})\cup \mathrm{cl}(T_{-})$ by step 3.3; conversely a nonzero $v$ in that union satisfies $\mid v^0\mid \ge \mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}\mid$, hence $\eta (v,v)=-(v^0{)}^2+\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}{\mid }^2\le 0$ by step 1.1, and is therefore timelike or null, that is, causal, by [L3].

Let $(e_0^{\mathrm{\prime }},e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$ be another $\eta$-orthonormal ordered basis with coordinate map $c^{\mathrm{\prime }}$ and distance $d^{\mathrm{\prime }}(u,w):=\parallel c^{\mathrm{\prime }}(u)-c^{\mathrm{\prime }}(w){\parallel }_2$; then $c^{\mathrm{\prime }}=Mc$ for the change-of-basis matrix $M$ between the two ordered bases, which is invertible because both are bases [L2]. For any real matrix $N$ put $K(N):=({\sum }_a({\sum }_b\mid N_{ab}\mid {)}^2{)}^{1\mathrm{/}2}$; each component of $Nx$ satisfies $\mid (Nx{)}_a\mid \le ({\max }_b\mid x_b\mid ){\sum }_b\mid N_{ab}\mid \le \parallel x{\parallel }_2{\sum }_b\mid N_{ab}\mid$ by [L5], so $\parallel Nx{\parallel }_2\le K(N)\parallel x{\parallel }_2$, and applying this to $M$ and to $M^{-1}$ with $\kappa :=\max (K(M),K(M^{-1}),1)\ge 1$ gives ${\kappa }^{-1}\parallel x{\parallel }_2\le \parallel Mx{\parallel }_2\le \kappa \parallel x{\parallel }_2$ for every $x$; hence $d^{\mathrm{\prime }}$ and $d$ determine the same open sets, the same closures and the same connected subsets, so the components of $T$ and the openness and closure claims of this lemma are intrinsic, independent of which $\eta$-orthonormal basis defines the coordinates. Since the argument of steps 1.1 through 4.3 used only that $(e_0^{\mathrm{\prime }},e_1^{\mathrm{\prime }},e_2^{\mathrm{\prime }},e_3^{\mathrm{\prime }})$ is $\eta$-orthonormal, it applies verbatim in the primed coordinates and shows that the primed coordinate sets $\{v:v^{\mathrm{\prime }0}>\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}^{\mathrm{\prime }}\mid \}$ and $\{v:v^{\mathrm{\prime }0}<-\mid v_{\mathrm{v}\mathrm{e}\mathrm{c}}^{\mathrm{\prime }}\mid \}$ are also exactly the two components of $T$; hence the primed pair coincides with the unordered pair $\{T_{+},T_{-}\}$. By [L15] a time orientation is a choice of one of these two components, so the two displayed coordinate sets coincide with the chosen $C^{+}$ and its complement $C^{-}$ in some order, possibly the reverse of a previously fixed naming; in particular the description is basis-independent up to that interchange.