4

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.

MSIS
  • 821

1 Answers1

1

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.

Paul Frost
  • 93,232
  • 2
    Actually one can talk about transversality in the topological setting. You say the two curves have finitely many intersections and that at each point of intersection there is a neighborhood and a homeomorphism of that neighborhood to a standard disk, carrying the two curves to two straight crossing lines. This can be developed further to higher dimensional manifolds. I first learned about it in Greenberg and Harper's text on Algebraic Topology where they use it to develop intersection theory for homology. – Cheerful Parsnip Nov 21 '19 at 01:25
  • @CheerfulParsnip Interesting, I didn't know that. But in fact it seems clear the one can transform a feature of transversal smooth submanifolds into a definition for topological submanifolds. Or pehaps better, to start with this definition and prove that is characterized in the smooth case by the "usual" conditions on tangent spaces. – Paul Frost Nov 21 '19 at 10:43
  • @PaulFrost: I specified that the number intersections be finite and, in asking for transversality I am asking that the curves actual cross and not just touch each other. – MSIS Nov 23 '19 at 01:50
  • 1
    @MSIS Yes, but my point was that your terminology suggests that you work in the smooth category. See https://en.wikipedia.org/wiki/Transversality_(mathematics) . As Cheerful Parsnip explained, there are interpretations also for topological submanifolds, but that is not standard. – Paul Frost Nov 23 '19 at 09:44