# Two topology questions

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?

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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

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 $g(x)$.
@t.b.: I may have mis-stricken the balance between 'being helpful' and 'doing their homework', but it wasn't tagged as homework. There's still plenty to do (e.g. the other cases in (1), the countability or otherwise of $A$ in (2)). –  Clive Newstead Sep 10 '12 at 13:53
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