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 ?
If we remove a countable family of simple non-closed curves , which are mutually disjoint , from the plane , then is the resulting space connected?
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general-topology
algebraic-topology
metric-spaces
continuity
connectedness
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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? – 2017-01-18
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0Do you mean **a** countable family of such maps? – 2017-01-18
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0@MoisheCohen : yes I did mean a countable family . Thanks – 2017-01-18
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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. – 2017-01-18
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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. – 2017-01-19
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0@DavidHartley: You are right, I did not notice that. – 2017-01-19
1 Answers
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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.
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0This is not countable. (It seems that the condition of being countable was edited into the question after you wrote this answer?) – 2017-02-24