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Let $K$ be an infinite compact subset of $\mathbb{C}$. Is it true that there exists a sequence $(f_n)_{n>0}$ of functions holomorphic in some neighborhood of $K$, such that the images $f_n(K)$ are pairwise non-homeomorphic? (Motivation for this question comes from: consider an element in a unital Banach algebra; how does the topological type of its spectrum change under holomorphic functional calculus?)

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Special case of interest to me: $K$ is a closed disk (more generally, a homeomorphic image of such). – user53153 Feb 11 '13 at 17:37
@5PM Can't you map $K$ to a suitably thin neighborhood of $[0,1]$ and apply the mapping you used for $[0,1]$? – JSchlather Feb 11 '13 at 20:07
@JacobSchlather Indeed, that works. Hm, does not that also work for any compact set with a nontrivial bounded component? At least if this component is at positive distance from the rest of the set. We can send all other components to points, map the remaining one to something that looks like a segment (Mergelyan's theorem should help, although interior is a bit of a problem). – user53153 Feb 11 '13 at 20:14
@5PM I don't know, there are all sort of weird things that come into play. I spent all this time in complex analysis studying conformal mappings, which is almost never what we want here. For instance if the non-trivial bounded component is not simply-connected is it clear how to map it to a neighborhood of a line segment? Certainly if its a simply connected bounded component isolated from the rest it should be doable. I suppose if $K$ is a connected compact set. Can we find a holomorphic mapping $f$ on a neigbhorhood $U$ such that $[0,1] \subset f(K)$ and $f(U)$ fits in a small nbd. – JSchlather Feb 11 '13 at 20:26

Here are a few thoughts, not really an answer.

I think you want to loosen the definition up a little bit, or maybe you don't. At any rate if you consider $K=\{0,1,1/2,1/3,\dots\}$ and fix an open neighborhood $U$ of $K$ which has $n$ connected components which meet $K$ non-trivially then if $g: U \rightarrow \mathbb C$ is holomorphic $g(K)$ is either finite of cardinality less than or equal to $n$ or infinite with one limit point. Let $U_0$ be the connected component of $U$ containing $0$ it follows that all but finitely many points of $K$ are contained in $U_0$. In particular $K_0=U_0 \cap K$ has a limit point in $U_0$ so if $g$ takes finitely many values on $U_0$ it is constant, hence $g(K_0)$ is either a single point or an infinite set with one limit point.

I don't really think this example is interesting to you, so you're probably better off considering a sequence of functions $f_n$ such that each is holomorphic in a neighborhood of $K$ (and maybe that's what you originally meant).

We can reduce to the case that $K$ is a perfect set fairly easily as well. Write $K=K^\prime \sqcup I$ with $I$ the isolated points of $K$. If $|I| < \infty$ then note that if $V \subset \mathbb C$ is compact and $x \in \mathbb C$ then either $x \in V$ or $d(x,V)>0$ so we see that if we can't find infinitely many up-to-homeomorphism distinct images of $K^\prime$ then $I$ won't help. In the case that $I$ is infinite we can find $f_n$ such that $|f_n(K)|=n$, by taking suitable neighborhoods.

That's about as far as I can make it, I was trying to figure out the answer to the question for a simple set such as $K=[0,1]$. But I don't have a good repertoire of poorly behaving holomorphic functions so I couldn't get anything more than a loop or a line segment. Oddly enough I think I can do it for the Cantor set, it should be similar to the case of infinitely many isolated points.

As a final though if $K$ is actually the spectrum of an element of a unital Banach algebra, then I imagine there are some restrictions we could place on $K$ to make the task easier.

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For $K=[0,1]$, the sequence of polynomials $p_n(z)=(z-1/2+i)^n$ does the job. The image is a curve that spirals toward the origin and then away from it (but still moving in the same direction, say clockwise). This creates a lot of self-intersections (but finitely many, of course). Their number tends to infinity as $n\to\infty$; thus, choosing a subsequence we get infinitely many noh-homeomorphic images. // For more general compact sets, one can get crazily behaved holomorphic functions from Mergelyan's theorem. – user53153 Feb 11 '13 at 6:10
@Jacob: There is no restriction on $K$: let $(z_n)_{n>0}$ be a dense sequence in $K$. Then the diagonal operator $A=diag(z_n)$ on the Hilbert space $\ell^2$, is exactly $K$. – Tryphon Tournesol Feb 11 '13 at 9:50
I meant: has spectrum exactly $K$. Sorry. – Tryphon Tournesol Feb 11 '13 at 10:39

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