1

let $f:S^1 \to \Bbb{R}$ be a continuous map. Show there exists a point $x$ of $S^1$ such that $f(x)=f(-x)$.

My attempt: Let $g:S^1 \to \Bbb{R}$ defined by $g(x)=f(x)-f(-x)$, $\forall x\in S^1$. By theorem 21.5, $g$ is continuous. It’s easy to check, $-(-x)=x$ and $x \in S^1 \Rightarrow -x\in S^1$. By definition of order set, one and only one of the statements $f(x)\lt f(-x)$, $f(x)=f(-x)$, $f(x) \gt f(-x)$ is true. If $f(x)=f(-x)$, then we’re done. If $f(x)\lt f(-x)$, then $g(x)=f(x)-f(-x)\lt 0$ and $g(-x)=f(-x)-f(x) \gt 0$. So $g(x)\lt 0 \lt g(-x)$. By theorem 24.3, $\exists y \in S^1$ such that $g(y)=f(y)-f(-y)=0$. Thus $f(y)=f(-y)$. If $f(x)\gt f(-x)$, then $g(x)=f(x)-f(-x)\gt 0$ and $g(-x)=f(-x)-f(x)\lt 0$. So $g(-x)\lt 0\lt g(x)$. By theorem 24.3, $\exists y\in S^1$ such that $g(y)=0$. Thus $f(y)=f(-y)$.

Is this proof correct? I have some question, we used theorem 24.3 so we need to show $S^1 =\{x\in \Bbb{R}^2 | \| x\|=1 \}$ is connected. We can use Example 4 & 5, Section 24 of Munkres’ Topology proof technique, before that we need to show $\Bbb{R}^2 -\{0\}$ is connected. How to do that?

Note: For any given $x\in S^1$, we have constructed a point $y\in S^1$ such that $f(y)=f(-y)$. Which IMO is a different pattern of proof than exercise 3 section 24. In exercise 3 section 24, we essentially work with point $0$ and $1$.

311411
  • 3,537
  • 9
  • 19
  • 36
user264745
  • 4,143
  • 1
    Yes, it is correct. Recall that path-connectedness implies connectedness. It is easy to find continuous curves In $\Bbb R^2 - {0}$ that connect any two points. Most can be connected with a line. Those that can't can be connected with a polygonal curve with only two segments. – Paul Sinclair Mar 21 '22 at 13:32
  • @PaulSinclair opsss… – user264745 Mar 21 '22 at 15:23
  • 3
    Everyone fails to spot the obvious sometimes. Don't worry about it. – Paul Sinclair Mar 21 '22 at 16:10

0 Answers0