What is an example of an open map $(0,1) \to \mathbb{R}$ which is not continuous? Is it even possible for one to exist? What about in higher dimensions? The simplest example I've been able to think of is the map $e^{1/z}$ from $\mathbb{C}$ to $\mathbb{C}$ (filled in to be $0$ at $0$). There must be a simpler example, using the usual Euclidean topology, right?
Open maps which are not continuous
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real-analysis
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examples-counterexamples
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3Since $(0,1)$ and $\mathbb R$ are homeomorphic via a linear map composed with $\arctan$, it suffices to find a map $\mathbb R \to \mathbb R$ that is open but not continuous. Googling that gives you http://mathforum.org/library/drmath/view/62395.html – 2011-10-25
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1this is obviously not much help, but if you can find a continuous bijection $f$ with discontinuous inverse, then $f^{-1}$ will do. – 2011-10-25
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4One can build such a function from a Cantor set $C$ (the usual "middle thirds" set will do). Send each point in $C$ to $0$, and map each connected component of the complement of $C$ homeomorphically to the interval $(-1,1)$. Then the image of any open set intersecting $C$ will be $(-1,1)$ (thus open), and the image of any open set not meeting $C$ will also be open, since it's a union of homeomorphic images of open sets. Of course, each point of $C$ will be a discontinuity. – 2011-10-25
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2@PZZ for instance the map wrapping [0,1) around the unit circle. – 2011-10-25
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2@PZZ: In fact there are no counterexamples of the type you're suggesting: if $I$ and $J$ are intervals in $\mathbb{R}$ and $f: I \rightarrow J$ is a continuous bijection, then $f^{-1}$ is necessarily continuous. By coincidence this is exactly where I am in my Spivak calculus course, so see e.g. Theorem 37 in $\S 6.4$ of http://math.uga.edu/~pete/2400calc2.pdf. (Or see Spivak's text!) – 2011-10-25
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0@Pete, oh that is embarrassing. Since your reply is actually helpful, I will leave the comment where it is. – 2011-10-25