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Show that $ X = \{ (x,y) \in \mathbb{R}^2 : (x^2 +xy + y^2 -1)(x^2 + y^2 -8) = 0 \}$ is a non connected manifold of dimension $1$ in $\mathbb{R}^2$

First I can say that after I have shown that it is a manifold of dimension $1$, it is clear that it is not connected because then it would be of dimension $2$. But I have issues showing that it is a manifold because I know that it is wrong defining two functions $f(x,y) = x^2 +xy + y^2 -1$ and $g(x,y) = x^2 + y^2 -8$ and then taking the intersection. Because the intersection of two manifolds is not always a manifold.

vitalmath
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The set $X$ is the union (and not the intersection) of a circle and an ellipse. Since that circle and that ellipse do not intersect, and both of them are one-dimensional manifolds, their union is also a one-dimensional manifold.

José Carlos Santos
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  • How did you see so quick that the ellipse and the circle do not intersect? – vitalmath Nov 07 '21 at 17:44
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    Note that$$x^2+xy+y^2=\frac34(x+y)^2+\frac14(x-y)^2.$$Therefore, if $x^2+xy+y^2=1$, $|x+y|\leqslant\frac2{\sqrt3}$ and $|x-y|\leqslant2$. It follows that$$|x|=\frac12|(x+y)+(x-y)|\leqslant\frac1{\sqrt3}+1\quad\text{and}\quad|y|=\frac12|(x+y)-(x-y)|\leqslant\frac1{\sqrt3}+1.$$So,$$x^2+y^2\leqslant\frac83+\frac4{\sqrt3}<8.$$ – José Carlos Santos Nov 07 '21 at 17:58