[Math] continuity, geometrically

Question

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?

Answer 1

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.