Let $\mathcal F:=\{f:[0,1]\to \mathbb R^2 : f \mbox{ is injective }\}$ be a countable family . If $Im f \cap Im g=\emptyset , \forall f,g \in \mathcal F$ , then is it true that
$\mathbb R^2 \setminus \cup_{f \in \mathcal F} Im f$ is connected ?
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1Your set-builder notation of $\mathcal{F}$ means all injective maps from the interval to the plane. But the question in your title indicates $\mathcal{F}$ is just some set of injective maps. Which is it? – Matthew Leingang Jan 18 '17 at 16:38
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Do you mean a countable family of such maps? – Moishe Kohan Jan 18 '17 at 16:41
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@MoisheCohen : yes I did mean a countable family . Thanks – Jan 18 '17 at 17:05
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1The post http://math.stackexchange.com/questions/1321577/is-the-complement-of-countably-many-disjoint-closed-disks-path-connected?rq=1 contains a counter-example to the revised question. – Moishe Kohan Jan 18 '17 at 19:53
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1The example in 1321577 is connected and so does not answer this (revised) question. It only shows that the resulting space need not be path-connected. – David Hartley Jan 19 '17 at 10:01
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@DavidHartley: You are right, I did not notice that. – Moishe Kohan Jan 19 '17 at 16:33
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Any update on this? (So it does not need to be path-connected, but must it still be connected?) – Akiva Weinberger Oct 27 '22 at 15:47
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Let $f_x(t) = (x,t)$ for $x\in\mathbb{R}$ and $t\in[0,1]$. Then $f_x$ is an injective curve and $\operatorname{Im} f_x\cap \operatorname{Im} f_y \neq \emptyset$ if and only if $x=y$. The union $\bigcup_{x\in\mathbb{R}}f_x(t) = \mathbb{R} \times [0,1]$ separates the plane into two half planes.
Dan Rust
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This is not countable. (It seems that the condition of being countable was edited into the question after you wrote this answer?) – Akiva Weinberger Feb 24 '17 at 11:19