Let $\mathbb S^1$ denote the unit circle in $\mathbb R^2$.
Then prove that for every continuous function $f:\mathbb S^1 \to \mathbb R$, there exist uncountably many pairs of distinct points $x, y$ in $S^1$, such that $f(x)=f(y)$.
Real Analysis – Continuous Function on S¹ with Uncountably Many Distinct Points
general-topologyreal-analysis
Best Answer
Assume $f$ is non constant. Take a point $y_0$ between the minimum and maximum of $f$.
Assume the $f^{-1} (y_0)= \{x_0 \}$ i.e. it consists of only one point of $\mathbb{S}^1$.
Then $ f(\mathbb{S}^1/{x_0} )$ must be connected which leads to a contradictions.
So the preimage of $y_0$ has at least two points.