is the real line homeomorphic to the curve $\sqrt{x}$? I think no because of origin in $\sqrt{x}$. it's bounded on left but R is not bounded? is it just?
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1Real line for example is homeomorphic to open interval $(0,1)$ which is bounded. – RFZ Oct 25 '19 at 23:07
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1The curve $\sqrt{x}$ has a boundary point; the real line does not. That’s the real issue; you may be trying to say that, but “bounded” means something else. – Arturo Magidin Oct 25 '19 at 23:09
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@ArturoMagidin yes, I understand. Poor terminology from me sorry. so $\sqrt{x}$ without (0,0) is homeomorphic to real line? – Loli Oct 25 '19 at 23:13
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Somehow related question https://math.stackexchange.com/questions/240414/is-there-exist-a-homemoorphism-between-either-pair-of-0-1-0-1-0-1 – zwim Oct 25 '19 at 23:23
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@Loli Yes, that is right. – Milten Oct 26 '19 at 07:45
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Real line for example is homeomorphic to open interval $(0,1)$ which is bounded.
RFZ
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so we must remove point (0,0) from $\sqrt{x}$ then they are homeomorphic? – Loli Oct 25 '19 at 23:20
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$\frac 1 {\pi} (\pi /2+\arctan x)$ is a homeomorphism from $\mathbb R$ to $(0,1)$. $x \to \frac x {1-x}$ is a homeomorphism from $(0,1)$ to $(0,\infty)$. And $x \to (x,\sqrt x)$ homeomorphism from from $(0,\infty)$ to the graph of $\sqrt x$ if you restrict $x$ to $x>0$. Compose these maps to get a homeomorphism from $\mathbb R$ to graph of $\sqrt x$.
Kavi Rama Murthy
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