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Claim. Given an analytic map $f$ on a connected open set $U\subseteq\mathbb{C}$ and an open set $S$ such that $f(S)=\{0\}$, then $f=0$.

Proof. I define $Z=\{z\in U:f(z)=0\}$, then if I prove that $Z$ is both open and closed in $U$, it follows that $Z=U$ and we are done. I've already proved the closeness.

$Z$ is open: how to I proceed? My scattered thoughts

  1. I now that two analytic maps that agree on a set with accumulation point $z$, are equal on some neighborhood of $z$

  2. on one side if $Z$ is not open there is a point $z$ such that every open neighborhood meets $U-Z$; is this a contradiction? Not yet.

  3. At some point I have to use $S$. But the openness is a local thing (interior point)

  4. The connection between local and global could be done using a path (supposing that open subset of $\mathbb{C}$ is path-connected iff connected; or it is valid for general metric space, I miss some result here)

Q: How to I prove that $Z$ is open?

Note: At the end it is the Identity theorem, I think. But I would like to avoid the use of Taylor and derivatives. My feeling is that it should be possible with pure topological steps and using (1).

Mohamed Ali
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    See https://en.wikipedia.org/wiki/Identity_theorem – Kavi Rama Murthy Jan 28 '20 at 11:48
  • If you can use that two analytic maps that agree on a set with accumulation point are equal then it is trivial. Any open set has an accumulation point. – Conifold Jan 28 '20 at 11:54
  • @KaviRamaMurthy Your link is exactly the same as the one in my post – Mohamed Ali Jan 28 '20 at 12:11
  • @MohamedAli The identity theorem is also proved using connectedness, the way you are trying to prove. I don't think you can discover simpler proof. – Kavi Rama Murthy Jan 28 '20 at 12:14
  • No the two proofs are not the same! In the proof that actually works, instead of your $Z$ we havve the set $Z_\omega$, consisting of all the points where $f$ and every derivative of $f$ vanishes! (The proof that $Z_\omega$ is closed is the same as the proof that $Z$ is closed; the difference is it's easy to show $Z_\omega$ is open.) – David C. Ullrich Jan 28 '20 at 14:42

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Using $(1)$ a purely topological proof is possible:

Let $Z$ be the zero set of $f$ and $X$ be the set of all points in $D$, which are accumulation points of $Z$.

Since the limit of a convergent sequence of accumulation points of $Z$ is a accumulation point of $Z$, $X$ is closed in $D$. By $(1)$ (applied to $f$ and the the zero function), $X$ is open in $D$. Since $f\equiv 0$ on a nonempty open subset, $X\neq \emptyset$.

Hence by connectedness of $D$, $X=D$, so each point in $D$ is an accumulation point of the zero set of $f$. By continuity of $f$ (or again by $(1)$) $f\equiv 0$ on $D$.

Claire
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