Lower limit topology is a Hausdorff space $T_2$ Could you help me with the following?
Consider lower limit topology $\mathcal{T} $ with basis $\mathcal{B} = \{ [a,b) \ | \ a,b \in \mathbb{R}, \ a<b  \}$
Show that $\forall x,y \in \mathbb{R}, x\neq y \ \  \exists U,V \in \mathcal{T} \ : \ x\in U, y \in V, U \cap V = \emptyset$.
To be honest, I don't know what to use here. (I'm new to topology.)
 A: Hint: Assume without loss of generality $x < y$. You need to find one interval $I_1 = [a, b)$ in which $x$ lies, and one $I_2 = [c, d)$ in which $y$ lies, such that $a < b \leq c < d.$ These are all real number, so you can choose $I_1, I_2$ as...
A: You can also do this without actually choosing any intervals at all. Let $\mathcal{E}$ be the usual topology on $\Bbb R$. For any $a,b\in\Bbb R$ with $a<b$ we have 
$$(a,b)=\bigcup_{a<x<b}[x,b)\;,$$
so $(a,b)\in\mathcal{T}$. Every $U\in\mathcal{E}$ is a union of open intervals, so $\mathcal{E}\subseteq\mathcal{T}$. Now let $x$ and $y$ be any distinct real numbers. We know that $\langle\Bbb R,\mathcal{E}\rangle$ is Hausdorff, so there are disjoint $U,V\in\mathcal{E}$ with $x\in U$ and $y\in V$. But we just showed that $\mathcal{E}\subseteq\mathcal{T}$, so $U,V\in\mathcal{T}$, and $\langle\Bbb R,\mathcal{T}\rangle$ is therefore Hausdorff.
The more general principle here is that if $\tau$ and $\tau'$ are topologies on the same set $X$, $\langle X,\tau\rangle$ is Hausdorff, and $\tau\subseteq\tau'$, then $\langle X,\tau'\rangle$ is also Hausdorff: $\tau$ already contains enough open sets to provide each pair of points of $X$ with disjoint open nbhds.
