A cover $\mathcal C$ of a set $X$ by subsets of $X$ is called


$\bullet$ *minimal* if for every $C\in\mathcal C$ the family $\mathcal C\setminus\{C\}$ is not a cover of $X$;


$\bullet$ *minimizable* if $\mathcal C$ contains a minimal subcover of $X$. 


For example, any cover of the plane by parallel lines is minimal. 



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> **Problem.** Is each cover of the plane $\mathbb R^2$ by lines minimizable? What is the answer for the rational plane $\mathbb Q^2$?
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**Acknowledgement.** The problem was motivated by [this question](https://mathoverflow.net/questions/308265/minimal-covers-in-hypergraphs-with-finite-edges) of Dominic van der Zypen.




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**Added in Edit.** For the rational plane the affirmative answer can be also deduced from the following general



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> **Theorem.** A countable hypergraph $H=(V,\mathcal E)$ admits a minimal set $\mathcal M\subset \mathcal E$ with $\bigcup \mathcal M=\bigcup \mathcal E$ if each infinite set of vertices $I\subset V$ contains a finite set $F\subset I$ such that the set of edges $\{E\in \mathcal E:F\subset E\}$ is finite.
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The proof of this theorem is a bit complicated and will be presented in our joint paper with Dominic van der Zypen. 


We do not know if this theorem holds for arbitrary (not necessarily countable) hypergraphs.


On the other hand, @Peter Komjath in his comment to the answer of @bof claims that he can prove that every cover of the real plane by lines is minimizable, using some (difficult) general result on minimal covers of hypergraphs whose edges have the same cardinality and have small intersections. 



Is each cover of the plane by lines minimizable?
Unfortunately, you have two questions in one post. The one about $\mathbb R^2$ is too hard for me. The question about $\mathbb Q^2$ seems to have an easy affirmative answer, unless I'm making some dumb mistake. 


Let $P=\{p\_0,p\_1,p\_2,\dots\}$ be the set of points, and let $L$ be the set of lines. (In general, the construction seems to work for any countable bipartite graph which is $C\_4$-free and has no isolates.)


Let $L\_0=L.$ Consider the point $p\_0.$ Let $L'\_0$ be the set of all lines $\ell\in L\_0$ such that $p\_0\in\ell$ and $L\_0\setminus\{\ell\}$ covers $P.$ So $L\_0\setminus L'\_0$ covers at least $P\setminus\{p\_0\}.$ If $L\_0\setminus L'\_0$ covers $P,$ let $L\_1=L\_0\setminus L'\_0.$ Otherwise, choose some $\ell\in L'\_0$ and let $L\_1=(L\setminus L'\_0)\cup\{\ell\}.$


Continue in this way. I.e., Let $L'\_1$ be the set of all lines $\ell\in L\_1$ such that $p\_1\in\ell$ and $L\_1\setminus\{\ell\}$ covers $P,$ etc.


Finally, I claim that the set $L\_\infty=\bigcap\_{n=0}^\infty L\_n$ is a minimal cover.


Maybe the construction will be clearer in words. At step $n$ we have the set $L\_n$ of lines which have not yet been deleted (and this set still covers $P$), and we look at the point $p\_n.$ We look at the set of all (undeleted) lines through $p\_n.$ If such a line contains some point which is on no other (undeleted) line, we save it; if it contains no such point, we delete it. However, if this results in deleting *all* lines through $p\_n,$ then we save one of them.


Note that, after step $n,$ the point $p\_n$ is covered by at least one line $\ell\_n$ which will never be deleted, because it contains a point which is on no other line. Therefore $L\_\infty$ is a cover. It is a minimal cover because, if $\ell$ is in $L\_\infty,$ and if $p\_n\in\ell,$ then after step $n$ has been done there is some point on $\ell,$ either $p\_n$ or some other point, which is in no other surviving line.


**P.S** Now suppose $P=\mathbb R^2$ and $L$ is a cover of $\mathbb R^2$ by lines. I think the same construction will work **under some very restrictive assumptions** on the cover; namely, that we can enumerate the points as $\mathbb R^2=\{p\_\alpha:\alpha\lt\mathfrak c\}$ so that, for each $\alpha,$ either $|\{\ell\in L:p\_\alpha\in L\}|\lt\omega$ or else $|\{\ell\in L:p\_\alpha\in L\}|\gt|\alpha|.$ This is to prevent $p\_\alpha$ from getting uncovered at limit stages.


