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$x^3+y^3=\frac{1}{\sqrt{2}}$ is according to my textbook not bounded since, for each $x_0 \in R$ there is a $y_0$ such that the point $(x,y_0)$ is on the curve.

Isn't that true for a bounded set also e.g $x^2+y^2=1$. Im not sure what this statement says exactly.

F Wi
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Nope, that is not true, the difference between these 2 cases is that equation $y^3=a$ will have a real solution for any $a$, but $y^2=a$ will have a solution only for $a\geq0$, so if $|x|>1$ you will not find such $y_0$

AO1992
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  • got it! Also I am wondering how you can tell that x^3+y^2=1/sqrt(3) has points inside the circle that is all points such that: x^2+y^2 =< 1. – F Wi Aug 02 '19 at 10:31
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    @FWi Just consider some values of $-1\leq x\leq 1$ and plug them into $x^3 +y^3=\frac{1}{\sqrt{2}}$ after that check will your point be in a unit circle. For example, you can take $x=0$ – AO1992 Aug 02 '19 at 10:42