# If a function f has no jump discontinuities, does it have the intermediate value theorem property?

I realized I was confused by this concept (while preparing for my exam).

If a function f has no jump discontinuities, does it have the intermediate value theorem property?

Facts, I know: I know that continuity implies intermediate value theorem property. However, intermediate value theorem property does not imply continuity. All derivatives have the intermediate value theorem property, but we can have discontinuous derivatives. Derivatives do not have jump discontinuities, or discontinuities of the first kind.

I don't know if a function has no jump discontinuities then it necessarily has the intermediate value theorem property. I know this works for derivatives. I was trying to construct a discontinuous function with discontinuities of the second kind, (one of left or right does not exist), with no jump discontinuities, that doesn't have intermediate value theorem property, but I couldn't think of any.

Thanks

Consider $$f(x)=\begin{cases}\sin(1/x) & x>0\\ -5 & x\le 0\end{cases}$$

This has neither removable nor jump discontinuities, and yet never takes on the value $-4$.

No set $f:\mathbb{R}\to\mathbb{R}$ via $f(x)=x$ if $x\neq 0$ and $-1$ otherwise. Then $f(-1/2)=-1/2$ and $f(1/2)=1/2$ but there does not exist $x$ such that $f(x)=0$.

• Doesn't f have a jump discontinuity at 0? Both limits of x+ -> 0 and x- -> 0 exist and are 0. Feb 19, 2021 at 4:55

Another example: The characteristic function on the rationals:

$$\chi_{\mathbb{Q}} = \cases{1 & x \in \mathbb{Q} \cr 0 & x \notin \mathbb{Q}}$$

At no point does either one sided limit exist (so no jump discontinuities), but the function assumes no value between 1 and 0.