Is the set of points or the set of lines on a plane "larger"?

Is the set of points or the set of lines on a plane "larger",or there is a 1-1 correspondence between lines and points?

Solution 1:

Let's start building up a way of describing the set of all lines in the plane. There are two types of lines: vertical lines and non-vertical lines.

• Any vertical line can be represented by an equation of the form $x=c$, for some unique constant $c \in \mathbb R$. So if we call $S$ the set of all vertical lines, there is a clear one-to-one correspondence between $S$ and $\mathbb R$.
• Any non-vertical line can be represented by an equation of the form $y=mx+b$, where $y,b$ are unique constants in $\mathbb R$. So if we call $T$ the set of all non-vertical lines, there is a clear one-to-one correspondence between $T$ and $\mathbb R \times \mathbb R$.

The set of all lines in the plane is $S \cup T$, and the set of all points in the plane is naturally identified with $\mathbb R \times \mathbb R$. Now, what do you know about a union of two infinite sets? What do you know about the Cartesian product of two infinite sets? With that information you should be able to put the pieces together and answer the question.

Solution 2:

I assume the question asked by Marios was about the real plane, but planes over finite fields might also lead to interesting statements. First, let's take an example. Let $P$ be the affine plane over the two-element field $\mathbb{F}_2$. There are 4 points in this plane, but there are 6 lines (i.e. one-dimensional affine spaces) in this plane.

The proof for $\mathbb{R}$

We have a one-to-one mapping from $\mathbb{R}^2 \setminus\{(0,0)\}$ to $\mathcal{L}$ (I couldn't find simpler) : take any element $(a,b)$ and form the line $d_{a,b}$ which is tangent at the circle of center $0$ and radius $\sqrt{a^2 + b^2}$ at the point $(a,b)$. This is injective. It is not surjective, as the lines going through $(0,0)$ (let's denote it by $\mathbb{P}_1(\mathbb{R})$ ) do not have the form $d_{a,b}$.

Therefore, the set $\mathcal{L}$ have the same cardinality as $\mathbb{P}_1(\mathbb{R}) \cup \mathbb{R}^2 \setminus\{(0,0)\}$. But it is known that there is a bijection $\phi$ between $\mathbb{P}_1(\mathbb{R})$ and $\mathbb{R} \cup \infty$ : if $d \in \mathbb{P}_1(\mathbb{R})$ , let $\phi(d)$ be $x$, where $(x,1)$ is the intersection between $d$ and the horizontal line $y=1$. If $d$ is the line $y=0$, then $\phi(d) = \infty$. This is a bijection (it is clear on a drawing).

To conclude, we have a bijection of $\mathcal{L}$ onto $\mathbb{R}^3$. It can be shown that there is a bijection of $\mathbb{R}^3$ to $\mathbb{R}$.

The general case

In the general case, we might expect the number of lines to be "greater" than the plane.

First, note that there is a bijection of the projective space $\mathbb{P}_1(\Bbbk)$ onto $\Bbbk \cup \infty$, where $\infty$ is another point added to $\Bbbk$. When the field $\Bbbk$ is finite, it means that there is exactly as much one-dimensional vector spaces than elements in $\Bbbk$, plus one. When $\Bbbk$ is $\mathbb{R}$ or $\mathbb{C}$ they have the same cardinality.

Now note $\mathcal{L}_A$ the set of lines going through some point $x$ belonging to the set $A$. For instance, $\mathcal{L}_x$ is the set of lines going through $x$ and $\mathcal{L}_P$ is the set of all lines.

Note that $\mathcal{L}_x$ has the same cardinality as $\mathbb{P}_1(\Bbbk)$. Moreover, if $x \neq y$, then $\mathcal{L}_x \cap \mathcal{L}_y$ is a singleton (there is only one line going through $x$ and $y$ when $x \neq y$).

Using this, we show that the set of lines $\mathcal{L}$ is strictly greater than $P$, whatever be $\Bbbk$. In fact, take two points $x,y$ in $P$, we have a bijection of $\mathcal{L}_{\{x,y\}}$ onto $(\Bbbk)^2$ (we have removed an element counted twice).