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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?

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    @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?2011-01-20
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    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?2011-01-20
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    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.2011-01-20
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    @Theo Buehler: Thanks, I will try this.2011-01-20

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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.

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    +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 :)2011-01-20
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    @Theo: Thinking!2011-01-20
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    I gave a hint in my comments (Moreover, ...)2011-01-20