13b) Suppose $f$ is a function that satisfies the intermediate value theorem (IVT), and takes on each value only once. Prove that $f$ is continuous.
I looked up this question and found this proof on MSE:
The proof goes like this:
Suppose $f$ is not continuous.
$\therefore \forall \delta >0: \exists (\epsilon >0, x \in \mathbb{R}):\text{if } 0<|x-a|< \delta \implies |f(x)-f(a)|> \epsilon$
Annoying technicality 1) $|f(x)-f(a)|> \epsilon \implies f(x)<f(a) – \epsilon ~~~~\text{OR} ~~~~f(x)>f(a)+ \epsilon$
Annoying technicality 2) $0<|x-a| \implies x>a ~~~~\text{OR} ~~~~x<a$
Let's pick $f(x)>f(a)+ \epsilon$ and $x>a$.
Now the real proof:
(1) By IVT, $\exists y \in [a,x]: f(y)<f(a)+ \epsilon$
(2) By Assumption, $\exists z \in [a,y]:f(z)>f(a)+ \epsilon$
(3) By IVT, $\exists x_1 \in [z,y] \text{ and } \exists x_2 \in [y,x]:$
$f(x_1)=f(a) + \epsilon = f(x_2)
$
Which contradicts the fact that $f$ is injective.
I understand steps (2) and (3). It is step (1) that is causing me problems. The logic I see is that $f(x) > f(a) + \epsilon > f(a)$ so $\exists m:f(x)>m>f(a)$ which I can apply IVT to, to get: $\exists y \in [a,x]: f(x)>f(y)>f(a)$. But then how does it follow that $f(y)<f(a) + \epsilon$?
So then I thought what if I swap $f(x)$ with $f(a) + \epsilon$, but the problem then is I can't apply IVT since it only works for $f(x) < m < f(y)$ (the left and right sides of the inequality have to be a value of the function, which $f(a) + \epsilon$ is not necessarily).
What am I missing?
Edit: I figured it out. Trick is to apply IVT twice.
Best Answer
If $f(x)>f(a)+\epsilon>f(a)$, then $c=f(a)+\frac{\epsilon}{2}$ is a value between $f(x)$ and $f(a)$. Therefore, by IVT, there exists some $y$ between $a$ and $x$ such that $f(y)=c$. Now $f(y)<f(a)+\epsilon$.