0
$\begingroup$

How can I go about showing that if I have a function $f(x)$ that is continuous on an interval $I$, and there is a set $S$ where $f(x)$ fails to be 1 to 1. I need to show that $S$ is either uncountable or empty.

What is known:

$f(x) = f(y)$ for $y$ $\neq$ $x$ and $y$ $\in$ $I$ is the definition of one to one.

I know that if there exists an interval where 1-1 fails, I can use the proof that the subset of the reals is uncountable. However, what about singletons? Or the empty set?

$\endgroup$
  • $\begingroup$ Suppose $f(x) = x(1-x)$ on the interval $[0,1].$ Then $f$ is not 1-1 on the set $S=\{0,1\}. $ Yet $S$ has only two points. $\endgroup$ – zhw. Dec 8 '15 at 2:24
  • $\begingroup$ Any function is 1-1 on the empty set. $\endgroup$ – BrianO Dec 8 '15 at 2:28
  • $\begingroup$ I think what was wanted but mis-stated is that (1) the length of I is not 0, and (2) if f is not 1-to-1 on some non-empty S, then f is not 1-to-1 on some uncountable T. $\endgroup$ – DanielWainfleet Dec 8 '15 at 3:02
1
$\begingroup$

If $S$ is non-empty, then there exist at least two points $x,y$ such that $x < y$ and $f(x) = f(y)$. There are two possibilities:

  1. $f$ is constant on $[x,y]$, in which case $f$ fails to be one-to-one on a non-trivial interval.
  2. $f$ is not constant on $[x,y]$. Without loss of generality we assume there exists $\alpha \in (x,y)$ such that $f(\alpha) > f(x)$. By the intermediate value theorem, $f$ attains every value in $[f(x),f(\alpha)]$ at least twice, so $f$ fails to be one-to-one on an uncountable subset of $[x,y]$.

Exercise: show the conclusion of the statement can fail if we remove the continuity condition.

$\endgroup$
  • $\begingroup$ The question implicitly addressed the notion of singletons being potential answers, but can't $I$ be the empty set? Does that reduce to the trivial case where the function never fails to not be 1-1? $\endgroup$ – John Yates Dec 8 '15 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.