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Lets say that a set of lines of the real plane covers the plane if, for every element $\langle x,y\rangle\in\mathbb{R}^2$, there exists a line $l$ of the set that passes through $\langle x,y\rangle$.

Its obvious that the set of all lines of the plane, denoted by $\mathscr{L}$, has this property, and it can be identified with the real projective plane $\mathbb{P}_{\mathbb{R}}^2$ minus a point, say $(1:0:0)$. It is clear then that the cardinality of the set $\mathscr{L}$ is the same as the cardinality of $\mathbb{P}_{\mathbb{R}}^2$, but, how can we prove rigorously that every subset of $\mathscr{L}$ that covers the plane is actually equipotent with $\mathscr{L}$?

We know, for instance, that any line in $\mathbb{P}_{\mathbb{R}}^2$that doesn't pass through the distinguished point $(1:0:0)$, is a pincel of lines on the plane, and therefore, any line of $\mathbb{P}_{\mathbb{R}}^2$ that does not pass through $(1:0:0)$ can be identified with a set of lines that cover the plane.

I think this observation must be useful for the proof, but I don't know how to exploit this fact.

The thing is that any line of $\mathbb{P}_{\mathbb{R}}^2$ has the cardinality of $\mathbb{P}_{\mathbb{R}}^2$, and a pencil is one of the simplest subsets of $\mathscr{L}$ that cover the plane, because for every point $\langle x,y\rangle$ of the plane, there is only one line that passes through $\langle x,y\rangle$. "Intuitively", any subset of $\mathscr{L}$ that covers the plane must have a cardinal larger or equal to that of the pencil, because for a point $\langle x,y\rangle$ there might be more than one line that passes through it.

Since any pencil of lines through a point and the set $\mathscr{L}$ have the cardinality of $\mathbb{P}_{\mathbb{R}}^2$, from the previous commentary we should conclude that any two subsets of lines that cover the plane should have the same cardinality.

Hope someone helps me make a rigorous proof out of this intuitive and weak argumentation.

Thanks in advance for your time.

EDIT: just posted a full, complete answer to this question on the comments, after thinking a lot about this question. I hope that it is correct, and if not, I would really like to hear about your suggestions and comments.

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The set of all lines in the plane has the cardinality of the continuum.

If you have a family of lines covering the plane but which omits a line $L$, then there must be for each $P\in L$ a line in the family containing $P$. These lines must all be distinct, so there are continuum many of them.

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On the one hand, it is clear that the set of all lines on the plane, which we shall be denoted by $\mathscr{L}$, is a set of lines that cover the plane.

It is well-knon that this set can be identified with $\mathbb{P}_{\mathbb{R}}^2$ minut a point, say $(1:0:0)$, and then $|\mathscr{L}|=|\mathbb{P}_{\mathbb{R}}^2\setminus\{(1:0:0)\}|=|\mathbb{P}_{\mathbb{R}}^2|=|\mathbb{R}^2|=|\mathbb{R}|$.

Now, is $R$ is a set of lines that cover the plane, then we have that $|R|\le|\mathscr{L}|=|\mathbb{R}|$, because $R\subseteq \mathscr{L}$.

If $R=\mathscr{L}$, then there is nothing else to say about the cardinality of $R$, so we will suppose that $R\subset\mathscr{L}$. This implis that there exists a line $l$ such that $l\in\mathscr{L}\setminus R$.

We define in $R$ the following relation:

$$r\sim r'\Longleftrightarrow r\text{ and }r'\text{ meet in a point of the line }l,\text{ or both are parallel to }l$$.

Clearly, the relation $\sim$ is an equivalence relation, and we will consider the corresponding quotient set $R/\sim$, which is a partition of $R$.

We define the function $f:l\longrightarrow R/\sim$ defined by: for each $\langle x,y\rangle\in l,\;f(\langle x,y\rangle)=\{r\in R\,|\,\langle x,y\rangle\in r\}$, which clearly is a set, result of appling in $R$ the axiom of subsets corresponding to the property "$\langle x,y\rangle\in r$".

On the one hand, $f$ is well defined; by hypothesis, the image by $f$ of every element of $l$ is a nonempty set, and it corresponds to a unique equivalence class of $R/\sim$, because two distinct lines cannot meet in more than one point.

Moreover, $f$ is injective, for if $\langle x,y\rangle,\,\langle x',y'\rangle\in l$ but $\langle x,y\rangle\not=\langle x',y'\rangle$, then $f(\langle x,y\rangle)\not=f(\langle x',y'\rangle)$, because the only line that passes through $\langle x,y\rangle$ and $\langle x',y'\rangle$ is $l\in\mathscr{L}\setminus R$.

On the other hand, the axiom of choice assures that there exists a choice function $h:R/\sim\;\longrightarrow\bigcup R/\sim$, and Im$(h)$ is a set of representatives for $R/\sim$, which we shall denote by $S$. Note that $h$ is clearly injective, because $R/\sim$ is a partition of the set $R$.

Finally, define the function $F:l\longrightarrow S$ defined by: for each $\langle x,y\rangle\in l,\;F(\langle x,y\rangle)=(h\,\circ\,f)(\langle x,y\rangle)$, whcih clearly is well defined, because $\text{Im}(f)\subseteq\text{dom}(h)$, and it is injective, for it is a composition of injective functions.

We conclude that there exists an injective functions from $l$ to a subset of $R$, so $|l|\le|R|$. But $l$ is a line in the plane, and therefore, it is equipotent with $\mathbb{R}$, so $\mathbb{R}=|R|$.

By the Cantor-Bernstein theorem, $|R|=|\mathbb{R}|=2^{\aleph_0}$.

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