Proving if a function is continuous and not one-one then it has many such points.

Question

Let $g$ be a continuous function on an interval $A$ and let $F$ be the set of points where $g$ fails to be one-to-one; that is $$F = \{x \in A : f(x)=f(y) \text{ for some $y \neq x$ and $y \in A$} \}$$ Show that if $F$ is non-empty then it is uncountable.

An attempt at the solution.

Since F is non-empty there exists some $c_1,c_2$ $\in A$ such that $c_1\neq c_2 $ and $g(c_1)=g(c_2)$. Choose a point $z$ between $c_1$ and $c_2$ with $g(z) \neq g(c_1)$ Now $\forall$ $L$ satisfying $g(z)<L<g(c_1)$ or $g(z)>L>g(c_1)$ by $IVT$ there will be a point $K_1$ $\in (z,c_1)$ satisfying $g(K_1)=L$. Notice that $g(z)<L<g(c_2)$ or $g(z)>L>g(c_2)$ is also true for the same value of $L$ which by $IVT$ leads to the conclusion that there is also another point $K_2 \in (z,c_2)$ satisfying $g(K_2)=L$. Now how can i prove that there are uncountable number of points? Am i correct upto here?

Answer 1

You are almost correct, and almost there. If you say: let $z \in [c_1, c_2]$ s.t. $g(z) \neq g(c_1)$, you assume that such a point exists. You should remark that you can assume this without loss of generality, because if this is not true, then $g(z) = g(c_1)$ always holds so that $[c_1, c_2] \subset F$ and we are done.

You can then assume without loss of generality that $g(z) > g(c_1)$ (the other case is identical). You are right that for any $L$ strictly between $g(z)$ and $g(c_1)$, you can deduce by the IVT that there exist points strictly to the left of $z$ and strictly to the right of $z$ that are mapped to the same point. Since you can do this for any $L$, this actually implies that $[c_1, c_2] \backslash \{z\} \subset F$, so you are also done.