0

currently I am looking for a reference on the following topological statement. Consider, the space of curves with compact time interval on the whole complex plane given by $$ \mathcal{J} = \{\eta \, \colon [0,T] \to \mathbb C \mid \eta \text{ is continuous.}\} \qquad T > 0 \text{ fixed}. $$ Then the following statments should be equivalent for $\eta \in \mathcal{J}$,

  1. $\eta$ is simple.
  2. $\mathbb C \setminus \eta$ is connected.

Proof. We can see that (2) implies (1) by applying the Jordan curve theorem, i.e. if $\eta$ is non-simple then there exists some subset of $\mathbb C \setminus \eta$ that is a Jordan curve. As a result the complement $\mathbb C \setminus \eta$ must contain a curve. However, I am struggeling with (1) implies (2). If you know any reference, I would be very happy.

Thanks in advance!

a.s. graduate student
  • 840
  • 4
    The implication $2 \Rightarrow 1$ is false, and quite frankly I don't follow your argument at all. Look up space-filling curves. ...or even easier: Consider a constant curve, or one which is constant on a time interval, or... – Ben Steffan Mar 10 '25 at 17:25
  • 1
    The other implication is true though, see here for an argument and further links. – Izaak van Dongen Mar 10 '25 at 17:31
  • Thanks for the quick comments!

    @BenSteffan I am somehow still interested to modify this statement. By only considering curves that are "increasing", i.e. non-constant with some additional assumptions. If we take $(2)^{\prime}$ as: $\mathbb C \setminus \eta$ is connected, $\eta$ is non-constant, image of $\eta$ is compact and locally connected (this should already hold), do we get an equivalence?

     – a.s. graduate student Mar 10 '25 at 19:48
  • 1
    @a.s.graduatestudent That doesn't save you: Take any simple path and concatenate it with its path inverse. The implication $(2)' \Rightarrow (1)$ is also still false: At the risk of repeating myself, consider a space filling curve. You'll need to impose stricter conditions if you want a shot at equality: For a start, you'd do better to identify curves with the same image and ask when such a class has a simple representative, and then you'll also have to come up with something to avoid things like space-filling curves: Assuming your curves are smooth would accomplish that, for instance. – Ben Steffan Mar 10 '25 at 20:00
  • @Ben Steffan thanks, I understand the counterexample, just to clearify, you would take a space filling curve that covers, say a whole compact set $K$ as a counter-example? Somehow, I want to built the space of curves such that they are not like space filling curves that behave like "sandbags without holes". Anyways, maybe let me ask directly do you know of some topological property, lets call it $P$, such that $\mathbb C \setminus \eta$ enjoys $P$ implies that $\eta$ is a Jordan arc? – a.s. graduate student Mar 10 '25 at 20:16
  • 1
    @a.s.graduatestudent Sure. There are space-filling curves whose image is $[0, 1]^2$, say. As for a property, no, and I doubt you can find any. As I outlined in my last comment, working with equivalence classes of parametrizations and requiring smoothness would work. This is not a huge restriction: Every continuous curve is $\epsilon$-close to a smooth one for arbitrary choices of $\epsilon > 0$, and you can arrange the two curves to have the same endpoints and be homotopic relative to those, and the homotopy to also be through $\epsilon$-close curves (iirc), etc. – Ben Steffan Mar 10 '25 at 20:39
  • 1
    Anyways, this has strayed pretty far from the original question already. Comment sections are not for extended discussions; feel free to ask another question on the main site (after giving this some more thought yourself), or perhaps come to [chat]. – Ben Steffan Mar 10 '25 at 20:41

0 Answers0