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Here $X$ is defined as the set $\{(x,\frac{1}{x}\sin(1/x)) : x>0\} \cup\{ \text{y axis}\}$

Here is a sketch: Suppose there is such a $\gamma$ mapping $[0,1]$ to $X$, with $\gamma(0)=(0,0)$ and $\gamma(1)=(1,\sin(1))$. Then by IVT on the first coordinate, we have all points in $[0,1]$ being hit by the first coordinate. So construct sequences $x_n$, $y_n$ $\rightarrow 0 $ with $\sin(x_n)=0$, $\sin(y_n)=1$. We must have sequences $p_n, q_n \rightarrow 0$ with $\gamma(p_n)=(x_n,0)$ and $\gamma(q_n)=(y_n,1/y_n)$ ie no unique limit, contradicting continuity of $\gamma$

Edit: I'm pretty sure now it is not, and the flaw is in existence of sequences p and q.

beelal
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    Please recall the definition of X. – Colas Jun 06 '19 at 21:06
  • It might help to clarify the sequences $x_n,y_n$. You could probably use just one sequence something like $x_n = \frac{2}{n \pi}$, then examine how it behaves for odd and even $n$. – sfmiller940 Jun 06 '19 at 22:21
  • (1). You should have written $\gamma (q_n)=(y_n,1)$.... (2). The hypotheses imply that the sequences $p$ and $q$ do exist. What is missing is a demonstration that $p$ and $q$ must converge to $0.$ – DanielWainfleet Jun 07 '19 at 02:51
  • Could it be ammended as follows: first pick an $x_0$ with the desired property, then by IVT pick p_1 between $0$ and $p_0$ with gamma(p_1)=x_1... Then p_n bounded below decreasing so converges. Do the same for y. Then must be multivalued at 0 contradiction. Note p and q need not converge to 0, just be convergent. – beelal Jun 07 '19 at 07:26

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Your definition of the topologist's sine curve is somewhat unusual, but it is certainly a variant of it. The "standard" definition is $$T = \{(x,\sin(1/x)) : x>0\} \cup \left( \{ 0 \} \times [-1,1] \right)$$ which is compact.

The idea of your proof is correct, but it is not really clear how you define and use the sequences $(x_n), (y_n)$.

You assume that there is a map $\gamma : [0,1] \to X$ with $\gamma(0)=(0,0)$ and $\gamma(1)=(1,\sin(1))$ and invoke the IVT to show that $\gamma_1 = p_1 \circ \gamma : [0,1] \to [0,1]$ is surjective (where $p_1 : X \to [0,1]$ is the projection onto the first coordinate).

I suggest to proceed as follows. Define $x_n = (\pi/2 + 2n \pi)^{-1}$. Then $x_n \to 0$, $\sin (1/x_n)= 1$ and $\frac{1}{x_n} \sin(1/x_n) \to \infty$. Choose $t_n \in[0,1]$ such that $\gamma_1(t_n) = x_n$. The sequence $(t_n)$ must have a convergent subsequence $(t_{n_k})$ with limit $\tau \in [0,1]$. Since $\gamma$ is continuous, the sequence $(\gamma(t_{n_k}))$ converges to $\gamma(\tau)$. Because $\gamma_1(t_{n_k}) = x_{n_k} > 0$, we conclude that $\gamma(t_{n_k}) = (x_{n_k},\frac{1}{x_{n_k}} \sin(1/x_{n_k}))$. But this is a contradiction because $\frac{1}{x_{n_k}} \sin(1/x_{n_k}) \to \infty$.

Paul Frost
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