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1: Let $f\colon S^1 \to \mathbb{R}$ be any continuous map, where $S^1$ is the unit circle in the plane. Let:

$$A = \{(x, y) \in S^1 \times S^1 : x\ne y, f(x) = f(y)\}$$

Is $A$ non-empty? If the answer is ‘yes’, is it finite, countable or uncountable?

2: Let $f\colon S^1 \to \mathbb{R}$ be any continuous map, where $S^1$ is the unit circle in the plane. Let:

$$A = \{(x, y)\in S^1\times S^1: x = −y, f(x) = f(y)\}$$

Is A non-empty?

Please help anybody. I have no idea.

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For 2: A is not always empty. take $f(x)=c$ for all $x$. Same idea for 1. – Lucien Sep 10 '12 at 13:43
For 1) observe that $f$ attains its maximum and its minimum at two points $x_0,x_1 \in S^1$. This divides $S^1$ into two intervals. What can you say about their images? – t.b. Sep 10 '12 at 13:43
up vote 2 down vote accepted

We can identify $S^1$ with the interval $[0,1)$ via $e^{i\theta} \mapsto \dfrac{\theta}{2\pi} \pmod 1$, and so we can consider $f$ as a continuous function $[0,1] \to \mathbb{R}$ with $f(0)=f(1)$. We know that $f$ has to be bounded and attain its bounds, so let $x_*, x^* \in [0,1]$ with $f(x_*) \le f(x) \le f(x^*)$ for all $x \in [0,1]$.

Suppose $f(x^*) > f(0)=f(1)$. Then in particular $0<x^*<1$. By the intermediate value theorem, for each $0<t<1$, there exist $\xi_t \in (0,x^*)$ and $\eta_t \in (x^*,1)$ with $f(\xi_t) = t \cdot f(x^*) = f(\eta_t)$, and so $(\xi_t, \eta_t) \in A$. I'll let you consider the other cases; this answers (1).

For (2), consider the function $g : S^1 \to \mathbb{R}$ given by $g(x)=f(x)-f(-x)$. Prove that either $g \equiv 0$ or $g$ takes both positive and negative values. Hint: express $g(-x)$ in terms of $g(x)$.

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This exercice (2) admits an interesting generalization, naimly that for any continuous function $f:S^n\to\mathbb{R}$ ($n\geq1$), there exist antipodal points, i.e. such that $f(x)=f(-x)$. The proof is exactly the same. – Lucien Sep 10 '12 at 13:58
@Lucien, with more work you get the same result with codomain $\mathbb{R}^n$ as well (Borsuk-Ulam). (2) is the easy case $n=1$ of that theorem. – Henno Brandsma Sep 10 '12 at 14:15
@Henno Brandsma, thank you I did not know that! :-) – Lucien Sep 12 '12 at 10:31

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