Is a sphere a convex set?

**mamen** · November 27th 2009, 02:23 PM

Is it true that a sphere is a convex set?

Because the thing that I know is that the space enclosed by the sphere is the one convex.

thanks for the help.

**chiph588@** · November 27th 2009, 05:10 PM

A sphere is convex.

**Abu-Khalil** · November 27th 2009, 05:58 PM

In particular, a sphere encloses itself. Or you are asking bout $\Omega=\left\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=R ^2\right\}$ ?

**mamen** · November 28th 2009, 12:03 AM

Originally Posted by chiph588@

A sphere is convex.

But a sphere has a space inside where in that space is not a part of the sphere. So why is it convex?

**Opalg** · November 28th 2009, 12:41 AM

Originally Posted by mamen

Is it true that a sphere is a convex set?

Because the thing that I know is that the space enclosed by the sphere is the one convex.

The word "sphere" is used in two different senses. Sometimes it means a solid ball, and sometimes it means the surface of a ball. In mathematics, it's better to use the word "ball" for the solid ball, and to reserve the word "sphere" for the surface of the ball. In that case, the sphere is not convex. Of course, the ball is convex.

A similar thing happens in two dimensions, where the word "circle" is sometimes used to mean a disk, and sometimes (more accurately) it refers to the perimeter of the disk.

**HallsofIvy** · November 28th 2009, 04:00 AM

Yes, the "ball" is convex, the "sphere" isn't.

**mamen** · November 28th 2009, 02:13 PM

Originally Posted by HallsofIvy

Yes, the "ball" is convex, the "sphere" isn't.

yes I understand know that a sphere can either be convex or not, depending on the kind of sphere that we are refering to, right?

**HAL9000** · February 5th 2013, 12:39 AM

I would like to see how one shows that a sphere $S_r(x_0):=\{x \in E: ||x-x_0||=r\}$ in an arbitrary (not necessarily $\mathbb{R}^n$ ) normed space $E$ is non-convex.

I would say that this is simply due to the fact that for $x_1,x_2 \in S_r(x_0), t \in [0,1]$ the function $f(t):=||(1-t)(x_1-x_0)+t(x_2-x_0)||$ is non-constant on $(0,1)$ . But is it really an adequate argument? There seems to be too much of intuition coming from $\mathbb{R}^n$ in it, don't you find?

**HallsofIvy** · February 10th 2013, 05:59 AM

Originally Posted by mamen

yes I understand know that a sphere can either be convex or not, depending on the kind of sphere that we are refering to, right?

I wouldn't say "the kind of sphere". It's a matter of how rigorous one's language is. Using the word "sphere" [b]correctly[b] in mathematics, a sphere is NOT convex. If you use "sphere" to mean what should be called "ball", you are not speaking rigorously.

**HallsofIvy** · February 10th 2013, 06:02 AM

Originally Posted by HAL9000

I would like to see how one shows that a sphere $S_r(x_0):=\{x \in E: ||x-x_0||=r\}$ in an arbitrary (not necessarily $\mathbb{R}^n$ ) normed space $E$ is non-convex.

I would say that this is simply due to the fact that for $x_1,x_2 \in S_r(x_0), t \in [0,1]$ the function $f(t):=||(1-t)(x_1-x_0)+t(x_2-x_0)||$ is non-constant on $(0,1)$ . But is it really an adequate argument? There seems to be too much of intuition coming from $\mathbb{R}^n$ in it, don't you find?

I don't see any "intuition". The fact that f(t) is equal to r only for t= 0 or t= 1 follows from the triangle inequality in n dimensions.

**HAL9000** · February 24th 2013, 10:09 AM

I was actually talking about ANY normed space, possibly infinite-dimensional.

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