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Exercise 1.2.11(b) on page 21 of Jiří Lebl's SCV notes defines a complex line through the origin as the image of a linear function $L:\mathbb C\to\mathbb C^n$, and asks the reader to:

Prove that if an entire holomorphic function is bounded on countably many complex lines through the origin then it is constant.

It seems obvious that for $n>2$, $\mathbb C^n$ contains infinitely many distinct complex lines through the origin (e.g. if $L_n:\mathbb C\to\mathbb C^2$ is defined by $L_n(z)=(z,nz)$, consider the collection $\{L_n(\mathbb C)\}_{n\in\mathbb N}$.) Perhaps I'm misunderstanding complex lines, however, because I believe I have a proof, using the Lebl exercise above, that $\mathbb C^n$ cannot contain infinitely many distinct complex lines through the origin for any $n$:

Suppose for that sake of contradiction that for some $n$ there is a countably infinite collection of linear functions $\{L_j\}_{j\in\mathbb N}$, $L_j:\mathbb C^n\to\mathbb C$, such that the complex lines $L_j(\mathbb C)$ are all distinct. Then each function $L_j(z)=(a_{j,1} z,\cdots,a_{j,n} z)$ can be extended to a function $\widehat L_j:\mathbb C\to\mathbb C^{n+1}$ by defining $\widehat L_j(z)=(a_{j,1} z,\cdots,a_{j,n} z,0)$. This yields a countably infinite collection of distinct lines in $\mathbb C^{n+1}$, on which the function $f(z_1,\cdots,z_n,z_{n+1})=z_{n+1}$ is zero. Thus $f$ is a holomorphic function $\mathbb C^{n+1}\to\mathbb C$ which is constant on a countably infinite collection of complex lines, hence constant by Exercise 1.2.11(b). This is, of course, absurd.

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  • $\begingroup$ Related, but not identical: math.stackexchange.com/q/2641099 $\endgroup$ – A. Howells Feb 8 '18 at 21:27
  • $\begingroup$ For each $a=(a_1,a_2,...,a_n)\in\mathbb{C}^n$ the map $z\mapsto az=(a_1z,a_2z,...,a_nz)$ is a complex line. $\endgroup$ – orole Feb 8 '18 at 21:30
  • $\begingroup$ @orole Yes, but these aren't all distinct: The maps $z\mapsto az$ and $z\mapsto 2az$ have the same image, and thus should be considered the same line. Regardless, I'm pretty sure that there are infinitely many complex lines; the questions is really about finding the flaw with my "proof" to the contrary. $\endgroup$ – A. Howells Feb 8 '18 at 21:41
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    $\begingroup$ For me, your example works really fine and I think the exercise is just wrong. $\endgroup$ – Levent Feb 8 '18 at 21:44
  • $\begingroup$ Surely for each $v \in \mathbb{C}^n$ the function $z \mapsto zv$ is a line through the origin and there are uncountably many of these. $\endgroup$ – copper.hat Feb 8 '18 at 21:51
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Everything you say is correct, and the exercise is incorrect. The exercise should be stated only in the case that $n=2$. Or, if $n$ is allowed to be greater than $2$, then the function should be required to be bounded on infinitely many hyperplanes through the origin (linear subspaces of dimension $n-1$), instead of on infinitely many lines.

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