To begin, let me explain a proof strategy (which I'll call the connectedness principle for want of a better, more canonical term):

One way to prove that a mathematical object $O_1$ has some property $P$ is to:

  • Construct a (topological) space $\mathfrak M$ containing the object $O_1$ such that $P$ is a continuous invariant in $\mathfrak M$, meaning that if $t \mapsto O_t$ is a continuous path of elements of $\mathfrak M$, every element $O_t$ has property $P$ as soon as one of them has it.
  • Prove that some element $O_0$ in $\mathfrak M$ has property $P$.
  • Prove that $O_0$ and $O_1$ are linked by a path $t \mapsto O_t$ in $\mathfrak M$ (for example, prove that $\mathfrak M$ is path-connected).

This strategy is for example well illustrated by a very nice proof of the genus-degree formula.

My question is do you know any example of an application of this strategy in more elementary mathematics?

After all, even in high school or in the first years of college, the mathematical world is full of elements living in connected spaces (points, lines, triangles, numbers, functions...) and I see no reason that this proof strategy couldn't work in this context, even without the topology jargon.

Of course, one can imagine variants of this strategy (restraining oneself to polygonal paths, for instance) and I'm interested in all of these. However, I would like the answers to keep a topological flavour: I'm not interested (in this question) by examples using other kinds of invariance (e.g. through the action of a group).

A last remark: I haven't found this strategy in problem-solving books so I don't know if it has a well-established name. If you know of such a name, or of a book mentioning this strategy, please tell me!

  • $\begingroup$ define what you mean by elementary mathematics! $\endgroup$ – Mister Benjamin Dover Jan 11 '15 at 12:13
  • $\begingroup$ I understand that the vagueness of my question is frustrating (and I'm ready to apologise for that), but I am in any case unable to define "elementary mathematics". I guess it's one of those "I know it when I see it"-type things that are really hard to define... $\endgroup$ – PseudoNeo Jan 16 '15 at 19:48
  • $\begingroup$ Let me give an example of something that is not exactly what I'm looking for. I claim that when you draw the three medians of a triangle, the 6 triangles that are thus created (and which share the centroid as a vertex) have the same area. Proof: it's obvious for an equilateral triangle. I can pass from any triangle to an equilateral one via an affine transformation. Such a transformation does not change the property I want to prove (because an affine transformation multiply all the areas by the same factor – its determinant). $\endgroup$ – PseudoNeo Jan 16 '15 at 19:52
  • $\begingroup$ This is in my opinion a gorgeous proof. It shares one of the aspect I'm interested in: reducting a theorem to a trivial case, but the reduction is not the topological one I'm looking for. There is no a priori good reason that this equal-area property does not change when you move the triangle continuously (it would for example be obvious if we knew that all areas involved must be integers, but that's false). So my dream would be e.g. a property of triangles that you could prove saying: it's obvious for the equilateral triangle, it stays true when you deform a bit your triangle, so it's true! $\endgroup$ – PseudoNeo Jan 16 '15 at 20:08

I think you've misrepresented the first step in your proof strategy, because it's too strict and as such short-circuits all relationship to continuity (and you'll notice that your link does not use this method). Specifically, rather than proving that any continuous path preserves the property, you need only prove that the property is clopen, i.e. for any $x$ which satisfies $P(x)$ there is a neighborhood of $x$ which also satisfies it, and similarly for $\neg P(x)$. Furthermore, you need not have $\frak M$ path-connected, but only connected, because otherwise $\{x|P(x)\}$ and $\{x|\neg P(x)\}$ form a disconnection of the space.

This might be a bit too trivial, but the example that comes to mind for me is the proof of the intermediate value theorem. Let ${\frak M}=[0,1]$, $O_1=1$, $O_0=0$, and $$P(x)\iff f(x)<0\vee\exists y:f(y)=0.$$ (Note that the second disjunct does not depend on $x$.) Then if $f(x)$ is a continuous function on $[0,1]$ with $f(0)<0$ and $f(1)\ge 0$, we have $P(0)$ satisfied, and for each $x$, either $f(x)<0$ and then there is an open set $U$ around $x$ such that $f(y)<0$ for all $y\in U$, i.e. $P(x)$ is true on $U$, or the second disjunct is true so that $P(y)$ for all $y$. Otherwise $f(x)>0$ and the second disjunct is false, so $P(y)\iff f(y)<0$ for all $y$, which is false in a neighborhood of $x$. Thus the connectedness argument gives that $P(x)$ is true for all $x\in [0,1]$, and since $f(1)\ge0$ the first disjunct is false, so $\exists y:f(y)=0$.

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    $\begingroup$ I don't think I misrepresented the first step. In the degree-genus formula, it is clear that if the polynomial varies in the space of nonsingular polynomials (the space $\mathfrak M$), the corresponding (real) surfaces are deformations of each other and therefore they have the same genus. On the other hand, I agree that I was quite vague on the connected/pathwise connected stuff. To be honest, I hoped for examples where the topological part would be so obvious that this kind of distinction wouldn't matter. +1 for the second paragraph: it's not really what I am looking for... $\endgroup$ – PseudoNeo Jan 16 '15 at 19:41
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    $\begingroup$ but it's a nice spin on a classical proof. In any case, thank you for the time you spent on this question. I know it's frustratingly vague, and I sincerly appreciate the effort. $\endgroup$ – PseudoNeo Jan 16 '15 at 19:42

Well you can use it to prove that two homotopic contours in the plane minus a point have the same winding number about the point. This is done by integrating the function 1/z around the contours. Maybe this is not elementary enough.


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