Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
2  
Just because $\mathbb{R}$ is uncountable doesn't mean that it can't be Hausdorff. All you have to do to show a space is Hausdorff is pick any two points and put non-intersecting open sets around them. –  jmracek Jan 29 '13 at 15:26
add comment

3 Answers

up vote 2 down vote accepted

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).

share|improve this answer
add comment

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.

share|improve this answer
1  
To expand on this: You can show that the topology $\tau$ also contains the open sets $(a,b)$ for numbers $a < b$ by taking an increasing union of base sets. This should tell you that the identity function will be continuous. –  jmracek Jan 29 '13 at 15:31
add comment

(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 $\tau$?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.