Cfrgtkky

The orignal question is to prove that-"If $f$ has Intermediate Value Property and is monotonic then $f$ is continous"

I was looking for a proof and here is a proof by fellow mathematician.

An $epsilon,delta$-proof:

Let $x_0 in (a,b)$ be arbitrary and let $epsilon > 0$. Let $s_1 = min(f(x_0)+epsilon/2, f(b))$. Then the number $f(x_0)+s_1$ satisfies $f(x_0) < f(x_0)+s_1 < f(x_0)+epsilon$. By the intermediate value property, there exists then some $delta_1 > 0$ such that $f(x_0 + delta_1) = f(x_0)+s_1$ (by the choice of $s_1 = min(f(x_0)+epsilon/2, f(b))$ we are guaranteed to stay within the domain of $f$). Similarly, let $s_2 = max(f(x_0)-epsilon/2, f(a))$ and find a $delta_2>0$ such that $f(x_0-delta_2) = f(x_0)-s_2$. Take $delta = min(delta_1,delta_2) > 0$.

Now consider the punctured neighbourhood $U_delta = {x in (a,b) : 0 < |x - x_0| < delta}$. Without loss of generality, we may assume that $f$ is increasing. Let $x in U_delta$ be arbitrary. If $x > x_0$ we have, as $f$ is increasing, $f(x_0) leq f(x) leq f(x_0+delta_1) = f(x_0)+s_1 < f(x_0)+epsilon$, whence $|f(x)-f(x_0)| < epsilon$. If $x < x_0$, we have, as $f$ is increasing, $f(x_0)-epsilon < f(x_0)-s_2 = f(x_0-delta_2) leq f(x) leq f(x_0)$, whence $|f(x)-f(x_0)| < epsilon$.

You can work out the proof for the endpoints as an exercise!

I cannot understand why $f(x_0)+s_1$ satisfies $f(x_0) < f(x_0)+s_1 < f(x_0)+epsilon$, If this is correct then $s_1<epsilon$

if $f(x_0)+epsilon/2 < f(b)$ then $s_1=f(x_0)+epsilon/2 < epsilon$
Hence $f(x_0)<epsilon$

What is wrong here.

Monotone and IVP

The first paragraph as written is incorrect. Take $f(x)=x$ on $[0,1]$, $x_0=frac 12$ and $epsilon in (0,1/2)$. Then $f(x_0)+ s_1 = 1/2+1/2 + epsilon/2 > f(1)$.

Here is another approach. By monotonicity, the one-sided limits exist at all points, therefore only discontinuity is a jump discontinuity. A jump discontinuity contradicts the intermediate property.