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Let $S=\{x\in \mathbb{R}^n;|x|=1\}$ the unit sphere in Euclidean space, where $n>1$ and $|\cdot|:\mathbb{R}^n\rightarrow\mathbb{R}$ is a arbitrary norm.

The problem is to prove that $S$ is a infinit set. One way to prove it, is to show that there is a infinit set $X\subset \mathbb{R}^n$ such as $|x|=1$ for all $x\in X$. For example, $$X=\left \{\frac{(m,1,...,1)}{|(m,1,...,1)|}\in\mathbb{R}^n;m\in\mathbb{N}\right\}$$

Is there other way?

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Your set $X$ is not a subset of $\mathbb R$. Why don't you start by understanding why the unit circle in $\mathbb R^2$ has infinitely many points? –  Ted Shifrin Jun 19 '13 at 13:58
Actually, $X$ is a subset of $\mathbb{R}^n$. Sorry. –  Pedro Jun 19 '13 at 14:05
Your definition of $X$ still has an $\mathbb R$ in it. Also, $n$ is fixed, so use a different letter for the elements of $\mathbb N$. –  Ted Shifrin Jun 19 '13 at 14:12

2 Answers 2

up vote 0 down vote accepted

First, try to do it for $\|\cdot\|_2$:

$$f:\begin{array}{ll}[0,1]\to S\\t\mapsto \left(t,\sqrt{1-t^2},0,\dots,0\right)\end{array}$$

$f$ is injective since $\forall t_1,t_2 \in [0,1], f(t_1)=f(t_2) \implies \left(t_1,\sqrt{1-t_1^2},0,\dots,0\right) = \left(t_2,\sqrt{1-t_2^2},0,\dots,0\right) \implies t_1 = t_2$. This tells you that there are more elements in $S$ than in $[0,1]$. Since you know that $[0,1]$ is infinite, youcan therefore conclude that $S$ is.

But for another norm, this can't work since our function was made specifically for $\|\cdot\|_2$.

Let's start by taking $\alpha = \cfrac{1}{\|(1,0,\dots,0)\|}$. We have $\|(\alpha,0,\dots,0)\|=1$.

Now for any $x\in [0,\alpha],$ define

$$g_x:\begin{array}{ll}\Bbb R_+\to \Bbb R_+\\t\mapsto \left\|\left(x,t,0,\dots,0\right)\right\|\end{array}$$

$g_x(0)=\cfrac{x}{\alpha}\le 1$ and $\lim\limits_{t\to +\infty}g_x(t)=+\infty$ so by the mean value theorem, we have $t_0 \in \Bbb R_+,g_x(t_0)=1$. For any $x\in[0,\alpha]$, we'll call that $t_0$ $h(x)$.

Now define $$f:\begin{array}{ll}[0,1]\to S\\t\mapsto \left(t,h(t),0,\dots,0\right)\end{array}$$

It is properly defined since we chose the $h(t)$ so that the vector is in $S$ and it is still injective for the same reason as for $\|\cdot\|_2$. So $S$ is infinite.

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Try the 2-dimensional case and embed inductively circle of inferior dimension in "bigger" ones.

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