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I want to know if the following statement is true or not:

$\gamma : [a, b] \to \mathbb{R}^k$ is continuous then $E := \{ t \vert \exists s \neq t: \gamma (t) = \gamma (s) \}$ is closed.

So here, $E$ is just the set of 'crossing points' of the curve $\gamma$.

I hope this is true so that I can find 'the first' crossing point $t_0 = \min (E)$.

I found the statement is true for a special case where $\gamma$ is a polygonal path(finite number of consecutive line segments) but failed in this general case.

Edit:

The statement is simply false, as the counterexamples in the comments show. What I wanted to prove was the existence of $\min (E)$ and I succeeded with polygonal path since it had a simple structure: finite number of edges. My question was messed up while thinking about the general case.

FYI, My original problem was Show that any closed polygonal path can be decomposed into a finite union of simple closed polygonal paths, which is Exercise 8.7 of Joseph Bak, Complex Analysis. Here I wanted to use $\min (E)$ and got curious about the general case.

Calmadeas
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    What about the curve $t\mapsto\begin{cases}2t&t\leq1/2\2-2t&t\geq1/2\end{cases}$? Does $E=[0,1]\setminus{1/2}$, which is not closed? – Michael Burr Jun 18 '23 at 10:06
  • Perhaps you could share you proof of the polygonal approach. – Michael Burr Jun 18 '23 at 10:14
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    For a counterexample where $E$ is countable, consider $\gamma:[-1,1]\rightarrow\mathbb{R}^2$ defined by $$\gamma(t)=\begin{cases} (t,t\sin(\pi/t))& t\in(0,1]\ (-t,0)&t\in[-1,0]\end{cases}\$$We have that $E={1/n:n\in\mathbb{Z}\backslash{0}}$ which isn't closed. – Christian E. Ramirez Jun 18 '23 at 10:48
  • Keep in mind that the above counterexamples do not disprove your overall conjecture that $\min(E)$ exists. – Christian E. Ramirez Jun 18 '23 at 10:57
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    @MichaelBurr Oh.. Thank you for pointing out my silly mistake. Something is messed up. Maybe I should stick on my desired conjecture to solve the original problem following C-RAM's advice. – Calmadeas Jun 18 '23 at 12:18
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    @C-RAM This should be the counterexample I was looking for! Although I need to work with my problem from the beginning.. Thank you for your advice! – Calmadeas Jun 18 '23 at 12:21

1 Answers1

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Both the statement and the conjecture are false. I finally came up with a counterexample:

$ \gamma(t) = \begin{cases} 0 &t = 0 \\ \frac{1}{2} \left( \frac{1}{n} + \frac{1}{n + 1} \right) + \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n + 1} \right) \cos \left( 3 \left( \frac{1}{t} - n \right) \pi \right) &\frac{1}{n + 1} \le t \le \frac{1}{n}, n = 1, 2, \dots \end{cases} $

As $t$ changes from $1/n$ to $1/(n + 1)$, this function changes from $1/n$ to $1/(n + 1)$, then to $1/n$ and then to $1/(n + 1)$ again.

Thus $E = (0, 1]$ so that $\min (E)$ doesn't exist.

Calmadeas
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