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Consider lower limit topology [TEX] \mathcal{T}[/TEX] generated by basis [TEX] \mathcal{B} = \left\{ [a,b) \ | \ a,b \in \mathbb{R}, \ a<b\right\} [/TEX].

I've already proven that this topological space is Hausdorff space, meaning that for any [TEX]x,y \in \mathbb{R}, \ x \neq y[/TEX] we can find disjoint neighbourhoods

and that [TEX]\{ x \}[/TEX] is closed in this topology.

What I can't prove is that:

1) [TEX] \forall x \in \mathbb{R} \ \forall A \subset \mathbb{R} [/TEX] closed in [TEX](\mathbb{R}, \mathcal{T} ), \ x \not \in A \ \ \ \exists U, V \in \mathcal{T} \ : \ x \in U, \ A \subset V, \ U \cap V = \emptyset [/TEX] (I hope it's comprehensible)

2) Prove that each point in [TEX](\mathbb{R}[/TEX] has a countable neighbourhood basis in [TEX](\mathbb{R}, \mathcal{T})[/TEX]