How would I prove that if I have a function $f$ on $\mathbb R^1$, that is continuous and non constant, that its range is non countable?
Here's my thought.
Let $f$ be a continuous, non constant function on $\mathbb R^1$. Because $f$ is non constant, it is not connected. Therefore, there exists a subset of $\mathbb R^1$ that is both open and closed within the preimage of $f$. Therefore, either the domain or range of $f$ is disconnected. Thus there is not always an intermediate value on all intervals in $\mathbb R^1$…
Am I close at all?
Best Answer
Hint: let $a<b$ be two points so that $f(a) \neq f(b)$. Then either $f(a)<f(b)$ or $f(a)>f(b)$. Let's discuss the case $f(a)<f(b)$, and the other case can be handled similarly.
Let $c \in \mathbb R$ be any number in the range $f(a)<c<f(b)$. Note that there are uncountably many such $c$. Now what does the intermediate value theorem tell you?