Say $C_1, C_2$ are Jordan curves, i.e., simple-closed continuous curves in
the plane. I read a claim somewhere that , $C_1, C_2$ in general position (
meaning not tangent to each other) intersect each other an even number of
times (assuming, as @Prototank pointed out, they intersect each other
finitely-many times)
--including $0$
if they are disjoint. I think it is true, and this is my argument. Please
critique:
Per the Jordan Curve thm, a Jordan Curve divides the plane into two
disjoint regions, the interior and exterior of a curve.
Now, consider Jordan curves $C_1, C_2$ intersecting transversally.
Each time $C_1$ intersects $C_2$ , it goes from the exterior of $C_2$
to its interior and must eventually leave the interior and intersect
$C_2$ in the process. This is true for each intersection, so the number
of intersections must be even. Is this correct, or am I missing something?
Thanks.
Your question is not very precise. Your terminology "not tangent to each other" and "intersecting transversally" suggests that you consider smooth Jordan curves - only in that case it makes sense. If so, you actually restrict to compact smooth transversal $1$-dimensional submanifolds of $\mathbb R^2$. Their intersection is a compact $0$-dimensional submanifolds of $\mathbb R^2$, i.e. a finite discrete set. See for example https://folk.ntnu.no/gereonq/TMA4190V2018/TMA4190_Lecture12.pdf .
You may of course also consider arbitrary Jordan curves, but then you must explicitly require that they have only finitely many intersection points $x_i$ and that at these points the curves $C_1, C_2$ are"crossing" which means that for each $x_i$ and each open neigborhhod $U_i$ we find points of $C_2 \cap U_i$ in both components of $\mathbb R^2 \setminus C_1$.
This being said: Under these assumptions your proof is correct.