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Is the set $\{(x,y)\in\mathbb R^2:x^{2/3}+y^{2/3}\le1\}$ connected?

Please help me. I'm clueless.

Added: Is the set convex?

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    $\begingroup$ I can see it's not convex since $(1,0),(0,1)$ are in the set but $(.5,.5)$ is not. $\endgroup$ – Sriti Mallick Aug 4 '13 at 13:31
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    $\begingroup$ HINT: Show that it is star shaped. $\endgroup$ – asatzhh Aug 4 '13 at 13:34
  • $\begingroup$ That would be harder @asatzhh. $\endgroup$ – Sriti Mallick Aug 4 '13 at 13:38
  • $\begingroup$ No, star shaped is a great way to attack the problem. $\endgroup$ – davidlowryduda Aug 4 '13 at 13:41
  • $\begingroup$ What is $x^{2/3}$ when $x\leq0$? $\endgroup$ – Christian Blatter Aug 4 '13 at 15:02
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If $(x, y)$ is in the set, then so are the points $(kx, ky)$ for all $0 \le k \le 1$. That is, the line to the origin. So the set is connected.

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Hint: $B_{2/3}\stackrel{\rm{}def}{=}\{(x,y)\in\mathbb R^2\mid x^{2/3}+y^{2/3}\le 1\}$ is equivalently the set \begin{align*} B_{2/3} &= \{(x,y)\in\mathbb R^2\mid |x|^{2/3}+|y|^{2/3}\le 1\} = \{ {x}\in\mathbb{R}^2 \mid \lVert {x} \rVert_{2/3}^{2/3} \leq 1 \} \\ &= \{ {x}\in\mathbb{R}^2 \mid \lVert {x} \rVert_{2/3} \leq 1 \} \end{align*} that is the unit ball for a $p$-norm (where $p\stackrel{\rm{}def}{=}\frac{2}{3}$). This should settle the problem of whether it is connected.

PS: you can see what it looks like on this page.

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