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.

Let $\tau$ be the usual topology on the real line $\mathbb{R}$. Does there exists a topology $\tau_{0} \subset \tau$ such that $(\mathbb{R},\tau_{0})$ is homeomorphic to the figure eight? Also, is it possible to find a topology $\tau_{0} \subset \tau$ and a quotient space with this topology such that this space is homeomorphic to $\mathbb{R}$?

What's the trick for this one?

share|improve this question
    
@Theo Buehler: We can identify the two end points of 0 and 1 and to get a circle. But $\mathbb{S}^{1}$ is not homeomorphic to $\mathbb{R}$. Can you please explain a little bit more your hint? –  student Jan 20 '11 at 21:57
    
Decide that $0 \in \mathbb{R}$ should be mapped to the point with a crossing. How should you define the neighborhoods of $0$? Moreover, what do you know about quotient spaces of compact spaces? –  t.b. Jan 20 '11 at 21:58
    
In other words: choose $0$ as the crossing point and parameterize the figure eight by the reals in such a way that the ends of the real line approach $0$ again. –  t.b. Jan 20 '11 at 21:59
    
@Theo Buehler: Thanks, I will try this. –  student Jan 20 '11 at 22:11
add comment

1 Answer

We let $\tau_{0}$ be the subset of open sets of $\tau$ consisting of $U$ in $\tau$ so that if $U$ contains $0$ then $U$ contains both some interval $(a,\infty)$ and some interval $(-\infty,b)$. This essentially glues $\infty$ and $-\infty$ to $0$ giving a figure eight.

$0$ is the intersection of the circles, $\mathbb{R}^{+}$ is one branch and $\mathbb{R}^{-}$ is the other.

share|improve this answer
    
+1: That's precisely what I tried to explain (without strictly giving it away) in my comments. You haven't answered the second question, yet :) –  t.b. Jan 20 '11 at 22:26
    
@Theo: Thinking! –  anonymous Jan 20 '11 at 22:31
1  
I gave a hint in my comments (Moreover, ...) –  t.b. Jan 20 '11 at 22:32
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.