# Proving that any two sets of lines that cover the plane have the same cardinality

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.

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.

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, $$|R|=|\mathbb{R}|=2^{\aleph_0}$$.