Base for a topology, euclidean topology and Hausdorff

Question

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!

Answer 1

First note that not only is the collection $\mathbb{B}$ a base, every open interval $(a,b)$ is open in the base that is defined from this topology $\tau$ (this is called the Sorgenfrey line, usually): if $x \in (a,b)$ then $[x,b)$ is in $\mathbb{B}$ and contains $x$ and is a subset of $(a,b)$, so every point of $(a,b)$ is an interior point in $\tau$. This means that $\tau_{\epsilon}$, the Euclidean topology on $\mathbb{R}$, is a subset of $\tau$, as the former is generated by the open intervals.

This shows that the identity under (2) is continuous, but not a homeomorphism (it's not an open map, as all sets $[a,b)$ are not open in $\tau_{\epsilon}$). Also, as $\tau_{\epsilon}$ is Hausdorff we can already separate all points in $\mathbb{R}$ using open sets from that topology, which are all already in $\tau$, so the latter is also Hausdorff.

As all basic open sets of $\tau$ are in fact clopen (closed-and-open), regularity (and complete regularity as well) are trivial: if $x \notin C$, where $C$ is closed, some set of the form $O = [x,a)$ misses $C$ (these sets form a local base for $x$) and so $O$ and $\mathbb{R}\setminus O$ are disjoint open sets separating $x$ and $C$.

With some more work you can show that $(\mathbb{R},\tau)$ is hereditarily normal as well, first countable, Lindelöf and separable, but not second countable. It's a very commonly used example (see the Wikipedia entry as well).

Answer 2

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.

Answer 3

(a) Let $x<y$ be different points, then if $r<\frac{y-x}2$ then the intervals around $x$ and $y$ with radius $r$ are disjoint.

(b) Since, $\tau_\varepsilon$ can be generated by the open intervals $(a,b)$, this question is asking whether these remain open in $\tau$, and vice-versa, whether $[a,b)$ are open in $\tau_\varepsilon$. One of these is false.

(c) Hint: What are the closed sets according to $\tau$?