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Let $\alpha:\langle a,b\rangle \longrightarrow \mathbb{R}^2$ be a continuous function and injective

how to prove that $\alpha (\langle a,b \rangle )$ has empty interior ?

Any hints would be appreciated.

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    $\begingroup$ $\left<a,b\right>$ means closed interval between a and b? $\endgroup$ – Hanul Jeon May 4 '13 at 4:49
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    $\begingroup$ $\langle a , b \rangle $ is an open interval. $\endgroup$ – felipeuni May 4 '13 at 5:06
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Let $f:[0,1]\to\mathbb{R}^2$ be an injective and continuous map, then $f$ is a Jordan arc. By a immediate consequence of the Jordan curve Theorem the set $A=\mathbb{R}^2\setminus\{f([0,1])\}$ is connected and $\partial A=f([0,1])$. Since $A$ is open $\operatorname{int}(\partial A)=\emptyset$.

If $f:(a,b)\to\mathbb{R}^2$ is injective and continuous map we extend uniquely to a continuous function defined on $[a,b]$ but we can lose injectivity in a pair of points, for example in Jordan curves ($f(a)=f(b)$). Proof should be separated in several cases depending of injectivity of the extension and uses other consequences of JCT (JCT or variations can still be applied in these cases, and we know that the connected components of $f(I)^c$ are open sets with boundary $f(I)$).

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If $f^{-1}$ were continuous.

By contradiction, suppose the image has non-empty interior, then there exist a point $p\in I=(a,b)$ and $\epsilon>0$ such that $B(f(p),\epsilon)\subset f(I)$.

Consider $A=f^{-1}(B(f(p),\epsilon)\setminus\{f(p)\})$,

$B(f(p),\epsilon)\setminus\{f(p)\}$ is connected. Is $A$ connected?

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    $\begingroup$ Even if $C$ is connected and $f$ continous, $f^{-1}(C)$ need not be connected, and there is no reason that inverse $f^{-1}:f(I)\to I$ is continuous function. How to derive contradiction? $\endgroup$ – Hanul Jeon May 4 '13 at 5:19
  • $\begingroup$ @tetori Thanks, you're right. $\endgroup$ – Gaston Burrull May 5 '13 at 5:34

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