# Large class of curves which only intersect each other finitely many times

I am trying to find a large subset of piecewise-differentiable plane curves of finite length (subsets of $$\mathbb{R}^2$$) with the following property:

For any pair $$\gamma_1, \gamma_2$$ of curves in this class, their images $$\Gamma_1, \Gamma_2$$ are such that $$\Gamma_1\cap \Gamma_2$$ has finitely many connected components.

I have attempted to prove that this is the case for this set of curves:

Piecewise-smooth curves (finite length) in which each piecewise component is either a straight line segment or a curve whose derivative is injective,

but I have failed to produce a proof or a counterexample to the claim that this satisfies the desired properties.

Could anybody suggest how to prove it, or why they believe it may be a false claim? If it is false, would some further restriction produce the desired properties?

Obviously, one can restrict to looking at just one piecewise component at a time - and this way it is easy to show that no such curve can intersect a line segment at infinitely many points (using the injectivity of the curve). And of course, any line segment can only intersect another segment at one point or the entire segment. I have failed to produce a proof for when both components are curved. I believe that having infinitely many intersection points should lead to non-injectivity of the curves at some point, but haven't been able to show that.

This doesn't work. See the construction below. The best restriction I can come up with is a piecewise analytic curve. Then locally around any intersection point, both curves are the graphs of analytic functions, so their difference is analytic, so if it is $$0$$ infinitely often, it is simply the $$0$$ function. Thus the intersection number is locally finite, and by compactness of the curves, it is finite.
They will be the graphs of functions $$f_1,f_2$$ on the interval $$[0,1]$$ such that $$f_1(0)=f_2(0)=0$$, $$f_1'(0)=f_2'(0)=1$$. As long as $$f_i'(x)>0$$ everywhere, your derivative condition is satisfied. We will make it so that for every natural $$n$$, we have $$f_1(1/2^n)=f_2(1/2^n)$$ and $$f_1'(1/2^n)=f_2'(1/2^n)$$, but that they differ in the intervals in between. To do this, we let $$f_1''=0$$ and $$f_2''=g$$ and we want to pick some $$g$$ such that $$\int_{2^{-n}}^{2^{-(n-1)}}g(x)dx=0\\ \int_{2^{-n}}^{2^{-(n-1)}}xg(x)dx=0$$ To satisfy these, we pick some nonzero smooth function $$\varphi$$ compactly supported on $$[0,1]$$. On $$[0,3]$$, let $$\psi(x)=a\varphi(x)+b\varphi(x-1)+c\varphi(x-2)$$. By linear algebra, by picking $$a,b,c$$ appropriately, we can make them not all 0 and have $$\int_0^3 \psi(x)dx=\int_0^3 x\psi(x)dx=0$$ We then make $$g$$ on the interval $$[2^{-n},2^{-(n-1)}]$$ be a horizontal and vertical rescaling of $$\psi$$ with the vertical rescaling factor decreasing very fast with $$n$$ (I think $$1/n!$$ is enough, but I'm not sure). The fast decrease is to guarantee that $$g$$ is smooth at $$0$$. This completes the construction. You can also get uncountably many intersections (separated by points where they don't intersect) by using a similar construction based on a Cantor set.
• Why does having $f_i'>0$ imply the derivative condition? – Joe Previdi Nov 26 '19 at 21:57
• @MathTrain It doesn't. I meant having $f_{i}''>0$. Then construction can be fixed by adding some large constant to each $f_{i}''$, that is adding $Cx^2$ to $f_i$. – Sam Zbarsky Nov 27 '19 at 15:32