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Math Help - Is a sphere a convex set?

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    Post Is a sphere a convex set?

    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.
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    MHF Contributor chiph588@'s Avatar
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    A sphere is convex.
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    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\}?
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    Quote Originally Posted by chiph588@ View Post
    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?
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    Quote Originally Posted by mamen View Post
    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.
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    Yes, the "ball" is convex, the "sphere" isn't.
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    Quote Originally Posted by HallsofIvy View Post
    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?
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    Junior Member HAL9000's Avatar
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    Re: Is a sphere a convex set?

    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?
    Last edited by HAL9000; February 5th 2013 at 01:42 AM.
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    Re: Is a sphere a convex set?

    Quote Originally Posted by mamen View Post
    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.
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    Re: Is a sphere a convex set?

    Quote Originally Posted by HAL9000 View Post
    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.
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    Junior Member HAL9000's Avatar
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    Re: Is a sphere a convex set?

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