In their article "The roots of a polynomial vary continuously as a function of the coefficients" Gary Harris and Clyde Martin give a topological proof of the well-known theorem that the roots of a polynomial depend continuously on its coefficients. I don't understand the key moment of the proof of the "Treorem A" (stating that $\hat\sigma\colon\mathbb C^n/\sim\to\mathbb C^n$ is a homeomorphism).

First, the authors define the metric $d$ on $\mathbb C^n/\sim$ and prove that the topology induced by $d$ is the quotient topology with respect to $\pi\colon\mathbb C^n\to\mathbb C^n/\sim$. Then they show that $\hat\sigma|_{\overline{B(0,M)}}$ is a homeomorphism where $B(0,M)$ is the open ball about $0$ in $\mathbb C^n/\sim$ of radius $M$. Finally, they show that the map $\hat\sigma\colon\mathbb C^n/\sim\to\mathbb C^n$ is open. For this purpose the authors take an arbitrary open set $U\subset\mathbb C^n/\sim$, a point $x\in U$ and the open ball $B(x,\varepsilon)\subset U$ such that $\overline{B(x,\varepsilon)}\subset B(0,M)$ for some suitable values $\varepsilon$ and $M$. Then they write "Since $\hat\sigma$ is open on $B(0,M)$ it follows that $\hat\sigma(x)$ is the interior of $\hat\sigma(B(x,\varepsilon))$".

The last sentence I don't understand. By the above argument we know that $\hat\sigma|_{\overline{B(0,M)}}$ is a homeomorphis onto its image. Thus the image of any open set in $\overline{B(0,M)}$ is open in $\hat\sigma(\overline{B(0,M)})$. In particular the image of any open set in $B(0,M)$ is open in $\hat\sigma(B(0,M))$ but not necessarily in $\mathbb C^n$. It means that $\hat\sigma(B(x,\varepsilon))$ is a neighborhood of $x$ in $\hat\sigma(B(0,M))$. But why the set $\hat\sigma(B(x,\varepsilon))$ is open in $\mathbb C^n$? Why the authors conclude that the set $\hat\sigma(U)$ is open in $\mathbb C^n$? Thanks for helping.


Their proof is wrong and you identified the weak point. They published an addendum to plug the loophole ("In response to numerous queries ...") http://www.ams.org/journals/proc/1988-102-04/S0002-9939-1988-0934880-2/S0002-9939-1988-0934880-2.pdf A much better proof is given by Alexandrian, who also gives a nice result about real simple roots of polynomials with real coefficients http://users.ices.utexas.edu/~alen/articles/polyroots.pdf

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.