Is T strictly coarser than L? I'll provide my proof that RR is coarser than L and if someone could explain to me or let me know of a correct proof, that'd be great.
Let T = { V $\subset$ R $\vert$ if x $\in$ V, then there exists a $\in$ $\mathbb{R}$ such that x $\in$ [a , $\infty$) $\subset$ V}
L = { U $\subset$ R $\vert$ if x $\in$ U, then there exists a,b $\in$ $\mathbb{R}$ such that x $\in$ [a, b) $\subset$ U}
Claim. T $\subset$ L (proper subset of)
Proof. Let V $\in$ T and let x $\in$ V. Then there exists a $\in$ $\mathbb{R}$ such that x $\in$ [a, $\infty$) $\subset$ V. Then x $\in$ [a, x + 1) $\subset$ [a, $\infty$) $\subset$ V $\subset$ U $\in$ L, where a, x + 1 $\in$ $\mathbb{R}$. Then V $\in$ L.
Claim. L $\not$$\subset$ T (This is where I'm getting confused.)
Proof. Let U $\in$ L and x $\in$ U such that there exists a, b $\in$ $\mathbb{R}$, x $\in$ [a, b) $\subset$ U.
Why can't x $\in$ [a, b) $\subset$ [a, $\infty$) since if x is in an interval of real numbers a, b, can't it be in the interval of a to infinity. Like, if x$\in$[a,b), then x $\in$ [x, b) $\subset$ [x, $\infty$) $\subset$ V $\in$ T, where x $\in$ $\mathbb{R}$. I know that I say, let V$_{1}$ = [0, $\infty$) and say x = 1, then 1 $\not$$\in$ [0,1), would that be enough-- or can someone shed some light for me.
Also, from the book I'm studying from, it says the RR (Right Ray) is coarser than U (Usual Topology). I can see that an element of RR is an element of U, but I'm trying to find out why an element of U is not an element of RR, would it be like the counter example where if x = 1, then U$_{1}$=(0,2), then 1 $\not$$\in$ (2, $\infty$) . If someone could give me a concrete example I can understand, that'd be appreciated.
Consider the set $[0,1)$. Since $[0,1)\in L$ and $[0,1)\notin T$, $L\not\subset T$. We know that $[0,1)\notin T$ because otherwise $[0,1)$ would have to contain a set of the type $[a,+\infty)$.
Concerning the other question, you should post it as another question, explaining what the right ray topology is.