The intermediate value property

Question

I want to prove this statement, Assume $f:(a,b)\to \mathbb R$ has intermediate value property, then $f$ cannot have jump discontinuities.

So, i have two way to prove;

1. assume $f:(a,b)\to \mathbb R$ has intermediate value property, then either $f(a)<c<f(b)$ or $f(a)>c>f(b)$ there is $x_0\in(a,b)$ such that $f(x_0)=c$, we show to show that f is continuous on (a,b).

2. Use contradiction suppose that $f$ has jump discontinuities , so use assumption to find contradict.

Can someone give a detail for easy way to prove this .

Answer 1

Having the intermediate value property doesn't imply that $f$ is continuous in all points of its domain. Take, for example, the function $$f(x)=\left\{\begin{array}{rl}\sin\left(\frac{1}{x}\right)&\mbox{, if $x\neq0$} \\ 0&\mbox{, if $x=0$}\end{array}\right.$$ It has the intermediate value property but it is not continuous at $0$, since $\displaystyle\lim_{x\to0} \sin\left(\frac{1}{x}\right)$ doesn't exist.

You have to do the second approach. Suppose that $x_0$ is a point of jump discontinuity of $f$, that is, $$ \limsup_{x\to {x_0}_−}f(x)<\liminf_{x\to {x_0}_+}f(x). $$ Fix $\displaystyle c\in\left(\limsup_{x\to {x_0}_−}f(x), \liminf_{x\to {x_0}_+}f(x)\right)$. Since $\liminf_{x\to {x_0}_+}f(x)>c$, fix $\varepsilon>0$ such that $$ \liminf_{x\to {x_0}_+}f(x)-\varepsilon >c, $$ and, by $\liminf$ property, there exists $\delta_1>0$ such that $$ x\in(x_0,x_0+\delta_1) \Rightarrow f(x)\geq \liminf_{x\to {x_0}_+}f(x)-\varepsilon >c. $$ Analogously, we get $\delta_2>0$ such that $$ x\in(x_0-\delta_2,x_0) \Rightarrow f(x)<c. $$ Therefore, in the interval $(x_0-\delta_2,x_0+\delta_1)$, the only point that could be equal to $c$ is $x_0$. If $f$ had the intermediate value property, then we must have that $f(x_0)=c$. But notice that $c$ was an arbitrary value fixed in the interval $$ \left(\limsup_{x\to {x_0}_−}f(x),\liminf_{x\to {x_0}_+}f(x)\right), $$ so we could have fixed a value $c'\neq c$ and also conclude that $f(x_0)=c'$, which is a contradiction.