why $x=0 \notin S^1 ?$

Question

I have some confusion on this answer

Here is the outline of the solution

No. With the usual torus embedded in $\mathbb{R}^3$, lying on the $OXY$ plane, one has a natural projection onto that plane, $p:S^1×S^1\to \mathbb{R}^2$, which is continuous.

Two points on the torus have the same image if they are one above the other, in the same vertical line. In particular, they are in the same meridian of the torus, i.e. they have the same first coordinate. So, if $p(a,b) = p(c,d)$, $a = c$. This implies that the Borsuk-Ulam theorem fails on the torus because if $x=-x$, and then $x=0\notin S^1$.

My confusion : why $x=0 \notin S^1 ?$

My thinking : $S^1 =\{(x,y) \in \mathbb{R}^2 : x^2 +y^2 =1\}$

Now put $x=0$,then $\{(0,y) \in \mathbb{R}^2 : 0^2 +y^2 =1\} \implies y=1$

I think $x=0 \in S^1$

$x$ is a real number, not an element of $\mathbb R^2$. You showed that $(0,1) \in S^1$. — Héhéhé, Sep 04 '21 at 16:49
@КряжевАрсений that means $x=(0,0) $ Am i right ? — jasmine, Sep 04 '21 at 16:55

score 1 · Accepted Answer · answered Sep 04 '21 at 21:25

When you work in several dimensions, it is usual to denote by $x\in\mathbb R^d$ the vector instead of the first coordinate. For example writing in dimension $2$ $$ x = (x_1,x_2), \ \text{ and } \ 0 = (0,0). $$ Hence if $x=0$, then $|x| = 0 \neq 1$ so $x\notin \mathbb S^1$.

score 0 · Answer 2 · answered Mar 15 '22 at 11:13

This question is about an answer of mine, and while the accepted answer is correct, there is another interpretation, which is seeing $S^1$ as a subspace of the complex numbers. Since $\mathbb{C}$ and $\mathbb{R}^2$ are homeomorphic, there is no difference, topologically speaking, but now $0$ is literally the complex number $0$. This is actually the interpretation that I had in mind.

why $x=0 \notin S^1 ?$

2 Answers2