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?
Asked
Active
Viewed 161 times
7
-
1I can see it's not convex since $(1,0),(0,1)$ are in the set but $(.5,.5)$ is not. – Sriti Mallick Aug 04 '13 at 13:31
-
3HINT: Show that it is star shaped. – asatzhh Aug 04 '13 at 13:34
-
That would be harder @asatzhh. – Sriti Mallick Aug 04 '13 at 13:38
-
No, star shaped is a great way to attack the problem. – davidlowryduda Aug 04 '13 at 13:41
-
What is $x^{2/3}$ when $x\leq0$? – Christian Blatter Aug 04 '13 at 15:02
2 Answers
8
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.
Sriti Mallick
- 6,137
- 3
- 30
- 64
Joseph Quinsey
- 858
2
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.
Clement C.
- 67,323