# Is this an equivalent definition of continuous?

I have learned the definitions of continuity via $\epsilon, \delta$, via sequences and limits, and via topology, and all of them seem fine after studying them for a bit, but none of them jumped out at me at first as capturing the intuitive "you can draw the function without ever having to raise your hand" notion.

So I spent a little while trying to capture that notion, and this is the best I have:

A function $f : (a,b) \to \mathbb{R}$ is hand-continuous (on the interval $(a,b)$) if it satisfies the intermediate value theorem on any subinterval $[c,d] \subseteq (a,b)$. Here the IVT just means that if $f(c) \leq y \leq f(d)$ (or $f(d) \leq y \leq f(c)$) then there exists $x \in [c,d]$ such that $f(x)=y$.

To me, this captures the intuitive notion of being able to draw the function without ever raising your hand, so my question is, is a function being hand-continuous on $(a,b)$ equivalent to it being continuous on $(a,b)$?

The proof of the standard intermediate value theorem shows us that any continuous function is hand-continuous, so it remains to example the opposite direction.

I have been trying to fit this to the episilon-delta definition, but I have not had any luck so far, and I am starting to doubt whether or not the two are equivalent. Does anyone have any insight?

• Functions with the IVT property are sometimes known as Darboux functions. They don't necessarily have to be continuous. For a point discontinuity consider for example $\sin(1/x)$ at $0$. For a (pathological) example of a nowhere continuous Darboux function lookup the Conway base 13 function. – dxiv Nov 27 '16 at 5:32

Not every Darboux function is continuous. Consider for example $f(x) = \begin{cases} \sin(1/x) & x \ne 0 \\ 0 & x = 0\end{cases}$.