# How to prove Peano curve is surjective?

Peano curve is defined by iteration, while each step would split the tiny square in last step into 9, then connect the centres in a particular sequence.

By doing so, one might argue, that each segment at any step , has the two ends as rational numbers.

Then I’m in doubt here: does Peano curve pass through irrational numbers points, eg (1/π, 1/π)?

• Is that the space-filling curve, the one that passes through all points in a square? – Lord Shark the Unknown Jul 14 '18 at 7:09
• The image of the curve has two properties: 1. it is dense and 2. it is compact (hence it is closed). Combine these two informations to conclude. – Crostul Jul 14 '18 at 7:09
• @athos So is $(1/\pi,1/\pi)$ inside that square? – Lord Shark the Unknown Jul 14 '18 at 7:10
• @Crostul where can I find the proof of such two properties? – athos Jul 14 '18 at 7:10
• – Crostul Jul 14 '18 at 7:15

Perhaps part of your problem is that you are confusing the curves in the steps used in defining the Peano curve with the Peano curve itself.

Without going into the details, the construction defines a sequence of curves $P_n : I \to I \times I$, where $I = [0,1]$. The iteration you are talking about defines the curves $P_n$ for each $n$. But these are not the Peano curve. Instead, the Peano curve is defined by $$P(x) = \lim_{n \to \infty} P_n(x)$$ for all $x \in I$. While it is true that every point along the curves $P_n$ has at least one of its two coordindates rational, this does not apply their limit, as every irrational number is the limit of a sequence of rational numbers.

The key properties of the construction are

• for each $n$, the square $I \times I$ is divided up into $n^2$ smaller squares in such a way that the squares for $n+1$ are a division of the squares for $n$, and the size of these squares $\to 0$ as $n$ increases.
• If $A$ is an $n^{th}$ level square, then for all $m \ge n, P^{-1}_m(A)= P^{-1}_n(A)$, which is a closed interval in $I$.

By the first condition, if $p \in I\times I$ is a point, then for all $n$, it lies in some $n^{th}$ level square $A_n$, and because higher level squares partition the lower level squares, we have $A_m \subseteq A_n$ for $m \ge n$. Clearly, $p \in \bigcap_n A_n$, but since the size of the squares goes to $0$ as $n \to \infty$, the intersection cannot contain two points. Therefore $\bigcap_n A_n = \{p\}$.

Consider the closed intervals $P_n^{-1}(A_n)$. If $m > n$, then $A_m \subseteq A_n$ and so $$P_m^{1}(A_m) \subseteq P_m^{-1}(A_n) = P_n^{-1}(A_n)$$ Therefore, $P_n^{-1}(A_n)$ is a sequence of decreasing nested closed intervals within $I$, and so their intersection must contain at least one point $x_p$. I.e., $x_p$ is a value such that for all $n$, $P_n(x_p) \in A_n$. But since $p \in A_n$ and the size of $A_n \to 0$ as $n \to \infty$, the distance between $P_n(x_p)$ and $p$ must also go to $0$. That is, $$P(x_p) = \lim_{n\to\infty}P_n(x_p) = p$$

• Thanks a lot. You are quite right that I was confused that “the curves in the steps used in defining the Peano curve with the Peano curve itself“. I agree with the essential idea of your proof, however, I doubt on the 2nd property that “ If A is a n-th level square, then for all m≥n, $P^{-1}_m(A) = P^{-1}_n(A)$”, because the definition is not on the squares, but on the centres of the square — take $n=1$ as example, P defines a mapping from 9 points on I onto the 9 centres of the first level squares. But this is just a trifle detail. – athos Jul 16 '18 at 23:41