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I am trying to make a "challenge problem" for my (undergraduate) real analysis students. Currently, the students knows about connectedness, compactness in $\mathbb R^n$, functional limits and continuous functions from $\mathbb R^m$ to $\mathbb R^n$. They are also familiar with all the standard real analysis results on $\mathbb R$.

The goal for this problem is to show that the punctured plane is not simply connected. I set up the problem by defining the terms "path", "loop", "path homotopy", and "null-homotopic". But I could not come up with a proof accessible to real analysis students showing that the path parameterized by $\alpha(t) = (\cos(2 \pi t),\sin(2 \pi t))$ is NOT null-homotopic. Can someone help?

Things I definitely want to avoid: fundamental groups, Brouwer fixed point theorem, residue theorem.

Things I wish to avoid: There is a proof using Green's theorem, which I guess has the same flavor as the residue theorem in complex analysis. I think this is something students are able to understand. But since we have not talked about vector calculus in this course, it would be better if the proof I write down as solution does not involve Green's theorem.

P. Factor
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  • Think of co-area formula. To prove something generic about the maps from unit disk to the circle $S^1$. A map from unit disk is just a homotopy to the constant map, at the origin. But just a guess, not sure if it'll work. – Behnam Esmayli Apr 14 '17 at 04:18
  • Don't you mean you are trying to show that $\alpha(t)$ is NOT null-homotopic, i.e., can't be contracted to a constant path? – pre-kidney Apr 14 '17 at 04:29
  • @pre-kidney Yes I meant that. Thank you for pointing out! – P. Factor Apr 14 '17 at 04:33
  • Related question, with both answers including something you want to avoid: https://math.stackexchange.com/q/1654655/ – Jonas Meyer Apr 14 '17 at 04:37
  • I don't want to come across as rude, but what are you willing to accomplish with this? There are many "challenges" which could be ontopic for an introductory course of real analysis, and trying to prove that the circle is not simply connected is hardly one of them. – Aloizio Macedo Apr 14 '17 at 15:19
  • @AloizioMacedo Students have seen in class that $\mathbb R$ and $\mathbb R^2$ are different because $\mathbb R \backslash {0}$ is not connected while $\mathbb R^2 \backslash {0}$ is. I wanted to have an analogy between $\mathbb R^2$ and $\mathbb R^3$. But since I cannot find a proof elementary enough for the student, I have dropped this problem for something else. – P. Factor Apr 15 '17 at 17:12

1 Answers1

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You could invoke the winding number. For a curve $t\mapsto {\bf z}(t)$ in the right half plane you have the running polar angle $$\phi(t)=\arctan{y(t)\over x(t)}$$ and therefore $$\phi'(t)={x(t)y'(t)-y(t)x'(t)\over x^2(t)+y^2(t)}\ .$$ Your students will believe (or easily check) that this formula remains valid in the full punctured plane. It follows that the total polar angle increment $\Delta\phi$ along a closed curve $$\gamma:\quad t\mapsto {\bf z}(t)\ne{\bf 0}\qquad(a\leq t\leq b)$$ is given by $$\Delta\phi=\int_a^b \phi'(t)\>dt=\int_a^b {x(t)y'(t)-y(t)x'(t)\over x^2(t)+y^2(t)}\>dt\ .$$ (If you do not want to assume $\gamma$ smooth you can define $\Delta\phi$ using finite partitions of the interval $[a,b]$.) This $\Delta\phi$ has to be an integer multiple of $2\pi$, hence the winding number $$N(\gamma,{\bf 0}):={1\over2\pi}\int_a^b \phi'(t)\>dt=\int_a^b {x(t)y'(t)-y(t)x'(t)\over x^2(t)+y^2(t)}\>dt\tag{1}$$ is an integer.

As the RHS of $(1)$ is a continuous function of not shown "homotopy time" it follows that $N(\gamma,{\bf 0})$ is invariant under a homotopy. Now your curve $\alpha$ has $N(\alpha,{\bf 0})=1$, while a small circle $\gamma$ around the point ${\bf e}:=(1,0)$ obviously has $N(\gamma,{\bf e})=0$.

Christian Blatter
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