Base for a topology, euclidean topology and Hausdorff I am trying to solve an exercise from an old exam, but i am stuck at some parts of the exercise, which are not very clear for me or i can't move on. I would be very glad if someone could help me.
So, this is the exercise:
On $\mathbb{R}$ consider the collection $\mathbb{B} :=\left \{ [a,b) \subset \mathbb{R}: a,b \in \mathbb{R}, a<b \right \}$
(a) Prove that $\mathbb{B}$ is a base for a topology $\tau$ on $\mathbb{R}$ and that $\tau$ satisties the axiom $T_{2}$.
Ok, here I proved easily that this collection is a base. I don't understand why $\tau$ should be a topology of Hausdorff space...we know that $\mathbb{R}$ is uncountable and has infinitely many elements. In this case it can't be Hausdorff?
(b) Consider the identity function $I: (\mathbb{R},\tau )\rightarrow (\mathbb{R},\tau _{\varepsilon })$, where $\tau _{\varepsilon }$ denotes the usual euclidean topology. Is $I$ continuous? Is it an homeomorphism?
(c) Does $\tau$ satisfy the axiom $T_{3}$?
$T_{3}$ was about that a closed set and a point, which is not contained in this set, have disjoint open neighborhoods.
Thank you in advance! 

Let's understand the case of $T_2$. Consider some $x,y\in \Bbb R$ - we need to come up with two open neighborhoods, one per each point, that don't intersect. Let $\delta=|x-y|$ then
$$
  N_x = [x-\delta/3,x+\delta/3)\quad N_y = [y-\delta/3,y+\delta/3)
$$
which are clearly open do don't intersect. 
For the case b) - note that $\mathrm{id}$ being a homeomorphism implies that topologies are equivalent. This is clearly not the case since $[0,1)$ is not open in the Euclidean topology.
I hope these hints help you checking $T_3$ and a continuity of $\mathrm{id}$ by yourself, otherwise please tell what is unclear to you.
