# Is there a proof that any curve can be built from very small line segments?

It seems to me (who is quite the math novice) that a very important ‘statement’, for a lack of a better word, that is foundational to many mathematical topics is that a given curve, which is continuous and differentiable, can be built from a bunch of straight lines as long as we make those lines ‘small enough’. For a 2D case, I interpret this as being able to build a curve that traverses through a 2D plane by only using little $$\Delta x$$ ’s and little $$\Delta y$$ ’s. I am wondering how one goes about proving this statement. It seems to me a good starting point can be illustrated using the following picture: I suppose I should clarify that I am simply using this circle as a starting point for this argument...this could be any arbitrary curve (not just the circumference of a circle...though I suppose there is probably a proof out there that shows tiny sections of a curve can also be approximated by an arc length of a circle with a certain radius...but that's another question for a different time).

So the question I want an answer to is the following: As $$\Delta x$$ becomes very small (and its corresponding $$\Delta y$$, based on the behavior of the curve, or, more specifically, based on the function that describes the curve, also becomes very small ), does

$$(r*\Delta \theta) / (\sqrt{(\Delta y)^2+(\Delta x)^2)}$$ approach 1.0?

How would one go about proving this? I feel like most arguments that I can think of are rather circular…in that I have to use a property that is based off of what I want to prove in order to prove it! Is there a proof for this limit? Or is this just an axiom we accept to be true?

Edit 1: It has been brought to my attention that including the word "differentiable" as a characteristic of a curve creates a circular argument for what I would like to prove. The logic behind that claim is "if the curve is differentiable, then of course a curve can be decomposed into line segments because that is the definition of differentiable". Assuming this is true, please disregard the word 'differentiable'. I am interested in solving the previously referred to limit as if I never knew that calculus existed!

• Certainly, your equation holds for the circle. However, it seems like a space-filling curve might be a counterexample to the property you're trying to convey with the equation. – Peter Shor Jun 25 '19 at 23:49
• Hahaha, as I said in the introduction, I am not very experienced with math so I am not quite sure what that type of curve is. After looking at your hyperlink, I am not quite sure I can figure out why that is a counterexample. (I'm not quite sure my equation even 'has' a counterexample...as I am posing it as a question rather than a conclusion) – S.Cramer Jun 25 '19 at 23:52
• Are you requiring the curve to be differentiable or smooth (infinitely differentiable)? – eyeballfrog Jun 26 '19 at 0:48
• oh, i'm sorry. I thought that they were the same thing! I will make the edit now. I just meant differentiable (not necessarily infinitely) and continuous. – S.Cramer Jun 26 '19 at 1:02
• @S.Cramer: Careful! Indeed, there are certain curves that inherently have this property, but they are not necessarily differentiable. That's why my above comment says "includes but is not limited to". For instance, Lipschitz-continuous curves in a Riemannian manifold have the desired property but are not necessarily differentiable. Answering your question requires a solid understanding of at least real analysis (assuming that you are interested only in what happens in $\mathbb R^n$). Otherwise add some Riemannian or even metric geometry. – Alex M. Jun 28 '19 at 18:40