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Suppose I build a function such that its graph is a unique line that can be drawn without lifting the pen(everywhere or in a specific range.)

Is that function continuous?

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    $\begingroup$ yes it is and this is somehow the exact intuitive interpretation of continuity $\endgroup$
    – Surb
    Feb 28, 2015 at 18:44
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    $\begingroup$ What do you make of the Weirstrass function which is continuous everywhere but differentiable nowhere? It's not possible to visualize how a pen could draw the graph of this function. $\endgroup$
    – user4894
    Feb 28, 2015 at 18:51
  • $\begingroup$ Not necessarily, you haven't ruled out corners. $\endgroup$
    – user117644
    Feb 28, 2015 at 18:53
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    $\begingroup$ Technically it's a one line draw and is possible... $\endgroup$
    – Donna
    Feb 28, 2015 at 18:54

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Yes, but...it doesn't necessarily capture "all" continuous functions (mathematicians like to call this situation "sufficient, but not necessary"). The problem lies with "draw"-imagine a computer-controlled drafting apparatus. Even with our "best" technology, we cannot create an apparatus that draws just "a single point" (any mechanical/electronic apparatus has a certain "minimum pixel size", but mathematical points have "no size"). So there is a "physical limit" to the amount of "jerkiness/jaggedness" we can instruct this apparatus to carry out. Mathematically, however, we can imagine a curve (function) that is "jagged" at all levels (think: fractal) even though we have no "physical" means to draw it accurately (we can make a good approximation which can fool our eyes, though).

So mathematically, we go in a slightly different direction: we call a function "continuous" if two points "near" each other as inputs, remain "near" each other as outputs. The specifics of how this gets carried out then depend on what we mean by "near". ONE way (which is particularly well-suited for the real numbers) is to use the DISTANCE between two real numbers $d(x,y) = |x - y|$ as a "measure" (the technical term is METRIC) of "nearness", which leads to the standard "epsilon-delta" formulation of continuity: a function $f$ is continuous at $a$ if for any "neighborhood size" (especially the "small ones") ($\epsilon > 0$), we can find a "suitably small neighborhood" ("within $\delta > 0$") such that $d(x,a) < \delta$ means $d(f(x),f(a)) < \epsilon$; that is, points "near" $a$ wind up "near" $f(a)$ under $f$.

This clearly the case when we draw a curve with a pencil without "lifting it up" from the paper.

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Your intuitive definition is the starting point for a rigorous definition of continuity. But note that when your pencil draw a line this line is also ''smooth'', and this fact is the starting point for another definition. i.e. the definition of ''differentiable function''. Putting this two definitions in a right mathematical form we find that a line can be 'continuous but not smooth. See here:Is there only one continuous-everywhere non-differentiable funtion?

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    $\begingroup$ Another concept springing from drawing with a pencil is the idea of a rectifiable curve (a curve of finite length) - it's a little weaker than differentiability, but it's stronger than continuity. $\endgroup$ Feb 28, 2015 at 19:42
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    $\begingroup$ @Meelo: There are (pathological examples of) differentiable curves that are not rectifiable. $\endgroup$ Feb 28, 2015 at 23:57
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Is that function continuous ?

Yes, of course it is. But the real question is whether those are the only kind of continuous functions. And the answer to that depends on whether your pencil can travel at infinite speed over a finite interval. Otherwise, nowhere differentiable continuous functions, such as Weierstrass', would be excluded.

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