Is the unit sphere in a preHilbert space a total set?

Some context:

• Unit sphere: $S(0;1)=\{x\in\mathcal{H}|\sqrt{\langle x,x\rangle}=1\}=\{x\in\mathcal{H}|\langle x,x\rangle=1\}$
• $A\subset \mathcal{H}$ total (definition): if $y\in\mathcal{H} : \langle y,x\rangle=0, \forall x\in A$, $y=0$ (the only vector that is orthogonal to all the vectors of the set is the zero vector).

Now, I have to prove that the unit sphere is a total set. I have taken $y\in\mathcal{H}$ such that $\langle y,x\rangle=0, \forall x\in S(0;1)$, and, from here, I have to deduce that $y=\mathbf{0}$ is the only possibility... Here is my attempt: Suppose that exists $y\in\mathcal{H}$, $y\neq \mathbf{0}$ that satisfies the conditions that I have already mention. $||y||\neq 0$, as $y\neq \mathbf{0}$, so $\frac{y}{||y||}\in\mathcal{H}$, and in particular, as $\langle \frac{y}{||y||},\frac{y}{||y||}\rangle=\frac{1}{||y||^{2}}\langle y,y\rangle=\frac{||y||^{2}}{||y||^{2}}=1$, $\frac{y}{||y||}\in S(0;1)$; let's see what we have:

• $\langle \frac{y}{||y||},\frac{y}{||y||}\rangle=1$
• By hypothesis, as $\langle \frac{y}{||y||},\frac{y}{||y||}\rangle=\frac{1}{||y||}\langle y,\frac{y}{||y||}\rangle$ and $\frac{y}{||y||}\in S(0;1)$, we conclude that $\langle\frac{y}{||y||},\frac{y}{||y||}\rangle=0$

And this two form a contradiction that leads us to saying that the only possibility for $y$ is $\mathbf{0}$, and $S(0;1)$ is then, a total set.

I am not sure if this is a great argument... I someone could give me a hint if I am not going in the great direction, It would be highly appreciated!

Solution 1:

It is a lot of text, but I think the main idea is already present.

I would write it as follows:

Assume $x \perp S(0,1)$. If $x \ne 0$, then by assumption $x \perp \frac{x}{\|x\|}$. However, this already implies that $x=0$, because $0 = \langle x, \frac{x}{\|x\|}\rangle = \|x\| \implies x = 0$, a contradiction. Hence, the only vector that is perpendicular on $S(0,1)$ is the zero vector.