I was looking at the following proof in one of my older scripts, but I have some doubts and wanted to ask for a few clarifications here, first:
Let $A=\{x\in\Bbb R^n\mid \|x\|_2=1\}.$ Show that $\operatorname{Int}(A)=\emptyset.$
This is the proof written there:
Suppose $S\subseteq A$ is open. We are going to prove $S$ must be empty. For the sake of contradiction, let's assume $S\ne\emptyset,$ that is $\exists x=(x_1,\ldots,x_n)\in S.$ $S\subseteq A\implies \|x\|_2=1,$ that is $x_1^2+\cdots x_n^2=1.$ Now, because $S\subseteq A$ is open, $\exists r_1>0$ such that $B(x,r_1)\subseteq S.$ Now, consider $r_2=\min\{r_1,x_1\}$. Notice that $y=(x_1+r_2/2,x_2,\ldots,x_n)\in B(x,r_1),$ but $\|y\|_2^2=(x_1+r_2/2)^2+\ldots+x_n^2=1+r_2x_1+r_2^2.$ So, $\|y\|_2=1\iff r_2x_1+r_2^2=0\iff r_2(x_1+r_2)=0\iff [r_2>0]\iff x_1+r_2=0.$ If $x_1=0,$ the contradiction is obvious. Otherwise, define $r=\min\{r_2,x_1/2\}$. Then $\|(x_1+r_3,x_2,\ldots,x_n)\|_2\ne1$. In each case, we arrive at a contradiction, hence $K(x,r_1)\nsubseteq A.$
I don't think I quite understand the reasoning behind the step $r_2=\min\{r_1,x_1\}$ as we assume $x_1>0$ and don't seem to use $r_2\le x_1$ later on. Then, in the computation of $\|y\|_2,$ I think there should be $\|y\|_2^2=1+r_2x_1+\frac{r^2}4$ and then the whole further calculations collapse.
I would also like to prove $\operatorname{Int}(A)=\emptyset$ via contradiction and considering two antipodal points in the open ball $K(x,r_1)\subseteq S\subseteq A$ by concluding one of those antipodal points has to be further away from the origin, and hence, has a norm greater than $1$, that is it belongs to $\Bbb R\setminus \{x\in\Bbb R^n\mid \|x\|_2\ge 1\},$ but it seems a bit wavy. I also found a related question, but I wouldn't like to invoke the closure and the boundary in this task.
Is there anything I'm missing in the given proof or should it be improved?
