Callum Dudley

Answered

2022-07-03

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.

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.

Answer & Explanation

Alexis Fields

Expert

2022-07-04Added 14 answers

Consider

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

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

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

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

mistergoneo7

Expert

2022-07-05Added 3 answers

Another example: The characteristic function on the rationals:

${\chi}_{\mathbb{Q}}=\{\begin{array}{ll}1& {\textstyle \text{}x\in \mathbb{Q}\text{}}\\ 0& {\textstyle \text{}x\notin \mathbb{Q}}\end{array}$

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

${\chi}_{\mathbb{Q}}=\{\begin{array}{ll}1& {\textstyle \text{}x\in \mathbb{Q}\text{}}\\ 0& {\textstyle \text{}x\notin \mathbb{Q}}\end{array}$

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

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