# Finding the diameter of spheres in higher dimensions

I am currently trying to understand how we can calculate the diameter of a sphere in higher dimensions given its set notation. I understand intuitively that it is the farthest distance any two points on the sphere can be from each other, but I am having trouble expressing that in quantifiable terms.

For example, for the unit n-dimensional sphere $$S = \left \{ x\in \mathbb{R}^n:||x||\leq 1 \right \}$$, is there a general way to calculate its diameter? The way I've understood it is that the radius of the unit sphere is 1, and therefore the diameter is 2, but this feels wrong to me for higher dimensional spheres.

• The diameter is 2 - always twice the radius. You can prove that with the triangle inequality. Nov 23, 2022 at 15:45
• @Ethan Bolker Somehow I was not aware of the triangle inequality, this is useful, thanks! Nov 23, 2022 at 15:56
• Take an $n$-dimensional sphere. Choose a point on its surface. Say $P=(1,0,0,0,...)$. Then there is one point which has maximum distance, and that is $Q=(-1,0,0,0,....)$. Nov 23, 2022 at 15:59

The general way is known as the triangle inequality: for any pair of points $$x,y \in \mathbb{R}^n$$, $$\left\|x+y\right\| \leq \left\|x\right\| + \left\|y\right\|.$$ Vector-norms are homogeneous: $$\left\| \lambda \cdot x \right\| = |\lambda|\cdot\left\|x\right\|$$ for any $$\lambda \in \mathbb{R}^n$$. In particular $$\left\|-x\right\| = \left\|x\right\|$$, so we can rewrite the triangle inequality (by replacing $$y$$ with $$-y$$) as: $$\left\|x-y\right\| \leq \left\|x\right\| + \left\|y\right\|.$$ That is, the distance between two vectors is always bounded by the sum of their magnitudes. On the $$n$$-ball (which you call $$S$$, but in other places is called $$B$$) this means that for any pair of vectors $$x,y \in S$$, $$\text{dist}(x,y) = \left\|x-y\right\| \leq \left\|x\right\| + \left\|y\right\| \leq 1 + 1 = 2.$$ The diameter of a set is defined to be the smallest upper bound of the distance between any pair of points: $$\text{diam}(X) := \sup\left\{ \text{dist}(x,y) \mid x,y \in X\right\}.$$ In particular, $$\text{diam}(S) \leq 2$$ because of the earlier inequality.
On the other hand, given a point $$x \in S$$ such that $$\left\|x\right\| = 1$$, we can check that $$\text{dist}(x,-x) = 2$$, so $$2$$ is in fact the smallest upper bound on the set of distances: $$\text{diam}(S) = 2.$$
This argument works for any normed vector space, including stranger, infinite-dimensional ones, so the property that $$\text{diam}(S) = 2$$ is arguably more a consequence of the definition of a norm rather than the intrinsic geometry of Euclidean space.