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On a current problem sheet I've read the following: Let $G\subseteq\mathbb C$ be a domain and $f:G\to\mathbb C$ holomorphic. Let $(z_n)_n$ be a sequence with a limit point in $G$ and $f(z_n)=0$ for all $n$, then $f(z)\equiv 0$ for all $z\in G$ according to the identity theorem.

My question is this: Is this even true? The sequence $z_n=0$ for all $n\in\mathbb N$ has the limit point $0\in\mathbb C$ and $f(z)=z$ satisfies the condition, however $f$ is not equal to $0$ on $\mathbb C$. As far as I know the identity theorem needs limit points of sets to be applied. So the statement would be true if we said the following: Let $(z_n)$ be a sequence such that $\{z_n:n\in\mathbb N\}\subseteq G$ has a limit point in $G$. Now if $f(z_n)=0$ for all $n\in\mathbb N$, the identity theorem implies $f\equiv 0$ on $G$.

Am I correct?

RedLantern
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    $0$ is not a limit point of $z_n=0$ – Maja Blumenstein Jun 21 '19 at 11:54
  • The definition of a limit point:"Every neighborhood contains an element of the topological space other than the point itself" The constant sequence $0$ has a limit, but the concept of a limit point is different from that – Peter Melech Jun 21 '19 at 11:56
  • In all the definitions I found it is: Some $a\in\mathbb C$ is a limit point of a sequence $(z_n)$ if every neighborhood of $a$ contains infinitely many elements of the sequence. Only for limit points of sets it is required that these points differ from $a$. – RedLantern Jun 21 '19 at 11:57
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    infinitely many elements, not one element with infinitely many indices, then the definition is compatible to that of a limit point of sets. – Peter Melech Jun 21 '19 at 11:58
  • So the sequence $z_{2n}=1$ and $z_{2n+1}=-1$ has no limit points? That's nonsensical to me. – RedLantern Jun 21 '19 at 12:01
  • I see the problem, by the definition I learned these are limits of subsequences, but no limit points or accumulation points, may be this is of interest to You. – Peter Melech Jun 21 '19 at 12:09
  • Isn't there some common ground? Does the sequence $z_n=(-1)^n$ have limit points or not? – RedLantern Jun 21 '19 at 12:14
  • No limit points, but two isolated points. – Peter Melech Jun 21 '19 at 12:15
  • Well, I have two books on complex analysis here that say otherwise. Now I know what got me so confused... It's just a definition everybody does differently. – RedLantern Jun 21 '19 at 12:17
  • Yes, I read some confusing definitions too. I would suggest to make the difference:"every neighborhood contains elements ( in the sequence or set) other than the point itself: limit point" "there is an neighborhood that only contains the point itself: isolated point". Anyway good You pointed that out!+1 – Peter Melech Jun 21 '19 at 12:25
  • Thanks for clarifying though! If you are interested in the points associated with the green checkmark you can just post some random answer and I will reward it. – RedLantern Jun 21 '19 at 13:38

1 Answers1

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The discussion in the comments shows that there is no consensus about the concept of a limit point a of a sequence. This is not a new discussion. Probably the reason for dissenting opinions is that a limit point $p$ of a set $S$ definitely requires the existence of a point $q \in S \setminus \{ p \}$ in any neighborhood of $p$. See for example Limit point of sequence vs limit point of the set containing all point of the sequence, what is diffrernce between limit point of sequence and limit of sequence.

Thus the whole problem is a sort a linguistic problem.

My personal opinion is that RedLantern's interpretation is correct. See also https://www.quora.com/What-are-the-differences-between-limit-and-limit-point. I think the limit of a sequence (if it exists) deserves to be called a limit point of the sequence, and this applies to constant sequences.

Paul Frost
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