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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: 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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    I gave a hint in my comments (Moreover, ...)2011-01-20