# Two different analytic curves cannot intersect in infinitely many points

A curve in Euclidean space $\mathbb{R}^n$, $n \geq 2$ is $analytic$ if the coordinates of its points $x= x_{1},...,x_{n}$ can be expressed as analytic functions of a real parameter $x_{i}=x_{i}(t)$, $i=1,...,n$ and $\alpha \leq t \leq \beta$ and the derivatives $x'(t_{0})$ do not simultaneously vanish for any $t_{0} \in [\alpha, \beta]$. I search for a proof of the following fact:

If the set of intersection points of two analytic curves is infinite, then these curves coincide.

Can we prove the same as above if we relax the condition that the coordinates are analytic to the condition that the coordinates belong to the class $C^{\infty}$?

Edit: Thanks to below remark by Ramiro, to obtain the above implication, we have to assume that any two curves $K_{1}, K_{2}$ as in the question are such that $K_{1} \cap K_{2}$ is not another analytic curve.

2nd edit: As suggested Peter, we can reformulate our question as follows: Consider two immersed curves which are parameterized real analytically on compact intervals. If they have an infinite number of different intersection points, then their union is again a real analytic immersed curve.

Wrong: $t\mapsto \binom{t}{\sin(t)}$ and $t\mapsto\binom{t}{\cos(t)}$.

Edit: Grzegorz pointed out that $t$ is in a compact interval $[\alpha,\beta]$. Under this assumption the statement is correct.

Proof: Let $f,g:[\alpha,\beta]\to \mathbb R^n$ be the two real analytically parameterized curves which intersect in $f(t_i)=g(t_i), i=1,2,\dots$ different points. Then the $t_i$ have accumulation points in $[\alpha,\beta]$, thus $f$ and $g$ coincide on $[\alpha,\beta]$.

In the $C^\infty$ case the statement is wrong: Let $[\alpha,\beta]=[0,1]$ and consider $$f(t) = \binom{t}{e^{-1/t^2}\sin(1/t)},\qquad g(t)=\binom{t}{0}.$$

2nd edit: Noam pointed out that my proof above was not conclusive. So let me try again: Suppose that $f(t_i)=g(u_i)$ for sequences of distinct points. The $t_i$ have accumulation an point $t$ and the $u_i$ have an accumulation point $u$. By continuity, $x=f(t)=g(u)$. Now we change both parameterizations as follows: Choose a line $r\mapsto x + r.v$ for a vector $v$ such that both curves are transversal to the hyperplane $v^\bot$. Since both curves are immersions (if I understood the question right), such $v$ exists. Now consider the orthogonal projection of both curves onto this line. In a neighborhood of $r=0$ these are real analytic diffeomorphisms, so their inverses gives us new parameterizations of both curves in the parameter $r$. There are still infinitely many intersection points of the two curves near $x$, but these intersections happen now at the same parameter values $r_i$ with $r_i\to 0$ without loss. Now my proof from above applies.

I hope that I did not overlook something else this time. Many thanks for pointing out my mistakes.

3rd edit: I overlooked again something (Thanks SJR and Ramiro). So what I proved is: The intersection of the two curves contains an open interval in each curve. Note that each curve, as an immersion, is locally a real analytic submanifold.

4th edit: Grzegorz, to the new question you posed in the comment, the following comes to my mind: By a theorem of Whitney, any closed set in $\mathbb R^n$ is the zero locus of a smooth function. So take a set in $\mathbb R$ like the Cantor set which in uncountable, and a smooth function on $\mathbb R$ vanishing exactly on this set. The graph of this function and the $x$-axis are two $C^\infty$-curves which intersect in this set.

What do mean by: $K_1\cap K_2$ is not $C^\infty$?

5th edit: Grzegorz, instead of excluding Ramiro's example in your edit of the question, you could reformulate as follows:

Consider two immersed curves which are parameterized real analytically on compact intervals. If they have an infinite number of different intersection points, then their union is again a real analytic immersed curve.

• @Peter: Notice that $t \in [\alpha, \beta]$, a compact set. – Grzegorz Tomkowicz Feb 12 '13 at 17:06
• Wait, how do we know the intersection points have the same $t$-coordinates on both curves? In general we have $f(t_i) = g(u_i)$ for some $t_i,u_i$ in $[\alpha,\beta]$, and it's not obvious that $g$ can be analytically reparametrized to make each $u_i$ equal the corresponding $t_i$. – Noam D. Elkies Feb 12 '13 at 18:00
• Peter, are you overlooking the counterexample given in Ramiro's answer? – Sidney Raffer Feb 12 '13 at 22:15
• Peter, can you give an example of two curves $K_{1}, K_{2}$ of class $C^{\infty}$, as in the above question, such that $K_{1} \cap K_{2}$ does not belong to $C^{\infty}$ and $K_{1}$ and $K_{2}$ intersect in uncountably many points? – Grzegorz Tomkowicz Feb 14 '13 at 14:11
• Peter, by $K_{1} \cap K_{2}$ I mean that the intersection does not contain any open interval in each curve. I want to avoid the cases described below by Ramiro. So, your example works. – Grzegorz Tomkowicz Feb 14 '13 at 19:19

What about $t \mapsto (t,t)$ and $t \mapsto (t+1,t+1)$ for $t \in [0,2]$? They have infinite intersection but do not coincide.

On the positive side (following Peter´s answer and Noam´s comment), the set of $t$´s for which there is some $u$ such that $f(t)=g(u)$ is a finite union of points and intervals by the o-minimality of the structure involved.

Let $f\colon [0,1]\to \mathbf R^n$ and $g\colon[0,1]\to\mathbf R^n$ parametrize your curves. The set $C$ of $t\in[0,1]$ such that there exists $s\in[0,1]$ with $f(t)=g(s)$ is a closed subset of $[0,1]$. More importantly, it is definable in the sense of model theory in the language — called $\mathbf R_{\text{an}}$ — consisting of polynomial functions and so called restricted analytic functions, namely analytic functions defined over compact subsets of $\mathbf R^m$.

It is an important theorem (Gabrielov; Denef and van den Dries) that $\mathbf R_{\text{an}}$ is $o$-minimal. This means that definable subsets of the real line are finite unions of intervals.

From there, one should be able to analyse the situation further when the coincidence set is infinite. I would conjecture that it can only be the union of one or two intervals. Ramiro gave an example with one interval, though more complicated examples are possible, e.g., of the form $s\mapsto F(u(s))$, $s\mapsto F(v(s))$, where $u,v\colon[0,1]\to\mathbf R$ are analytic and $F\colon\mathbf R\to\mathbf R^n$ is a fixed curve. In particular, the set is a union of two intervals if $F$ is a parameterization of a circle and $u$ and $v$ are suitably chosen so as to draw two arcs which overlap twice.