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Prove that a subset of the real line which is topologically equivalent to an open interval is an open interval.

Suppose a subset $A$ is topologically equivalent to the open interval $B=(a,b)$. Then there exists a bijective mapping $f:A\rightarrow B$ such that both $f$ and $f^{-1}$ are continuous. Since $B$ is connected, and the map $f^{-1}$ is continuous, $A$ must also be connected. So $A$ must be an interval. It seems now that I only have to rule out the intervals that are closed at one end or both ends (I suppose an interval ending at $\infty$ or $-\infty$ are open.)

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You’re doing fine so far, and yes, the open rays $(\leftarrow,x)$ and $(x,\to)$ count as open intervals in this context, as does $\Bbb R$ itself. – Brian M. Scott Aug 29 '13 at 1:57
up vote 4 down vote accepted

To rule out intervals with one or both endpoints, you might note that every point of $(a,b)$ is a cut point of $(a,b)$: removing it leaves a space with two connected components. This is not the case when you remove, say, $c$ from $[c,d)$.

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I think I managed to come up with another argument: Suppose $f(c)=e\in(a,b)$. Then consider $x,y\in(c,d)$ such that $f(x)>e,f(y)<e$. By the intermediate value theorem, there exists $z\in(x,y)$ such that $f(z)=e$. But $f(c)=e$ also, and so $f$ is not a bijection. – Paul S. Aug 29 '13 at 2:51
But I still don't quite understand what it is about removing a point and leaving a space with two connected components. Why is that a problem? – Paul S. Aug 29 '13 at 2:51
@Paul: Yes, that argument works fine. As for the one that I suggested, note that if $x\in(a,b)$, then $(a,b)\setminus\{x\}$ is not connected and has $(a,x)$ and $(x,b)$ as connected components. Now suppose that $f:[c,d)\to(a,b)$ is continuous; $[c,d)\setminus\{c\}=(c,d)$ is connected, but its image under $f$ cannot be. – Brian M. Scott Aug 29 '13 at 2:55
That's a nice argument, Brian. Many thanks! – Paul S. Aug 29 '13 at 2:58
@Paul: You’re very welcome! – Brian M. Scott Aug 29 '13 at 3:05

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