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- Thread starter AxiomOfChoice
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Actually, I think I've got it (somebody please verify):

[tex]

x\in f(A) \Rightarrow f^{-1}(x) \in A \Rightarrow f(f^{-1}(x)) \in A \Rightarrow x\in A

[/tex]

[tex]

x\in f^{-1}(A) \Rightarrow f(x) \in A \Rightarrow x\in A

[/tex]

[tex]

x\in f(A) \Rightarrow f^{-1}(x) \in A \Rightarrow f(f^{-1}(x)) \in A \Rightarrow x\in A

[/tex]

[tex]

x\in f^{-1}(A) \Rightarrow f(x) \in A \Rightarrow x\in A

[/tex]

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Landau

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You might want to explain what f and A are. What are domain and codomain of f? A is subset of...?

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This is all geared toward showing that the Fatou set [tex]F[/tex] and Julia set [tex]J[/tex] of a rational function are completely invariant. Apparently, since [tex]F = J^c[/tex], showing that [tex]f(F) \subseteq F[/tex] and [tex]f(J) \subseteq J[/tex] amounts to showing [tex]f(F) \subseteq F[/tex] and [tex]f^{-1}(F) \subseteq F[/tex], which is apparently equivalent to showing [tex]z_0 \in F[/tex] iff [tex]f(z_0) \in F[/tex].You might want to explain what f and A are. What are domain and codomain of f? A is subset of...?

The above questions about general [tex]A[/tex] and general [tex]f[/tex] is part of my attempt to understand these equivalences.

I should note that "apparently" is a stand-in for "according to Gamelin's

- #5

Landau

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A general function f still has a domain and a codomain. For expressions like [tex]f(A)\subseteq A[/tex] to make sense, A still has to be a subset of the domain of f, and furthermore A has to be a subset of the codomain of f. That's all information you didn't provide; so: what is f and what is A?

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Ok...well, [tex]A = \mathbb{C}^*[/tex], and [tex]f: \mathbb C^* \to \mathbb C^*[/tex] is a rational function. Does that help?

A general function f still has a domain and a codomain. For expressions like [tex]f(A)\subseteq A[/tex] to make sense, A still has to be a subset of the domain of f, and furthermore A has to be a subset of the codomain of f. That's all information you didn't provide; so: what is f and what is A?

- #7

Landau

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* f(A) is just the image of f, and by definition the image of f is a subset of the codomain.

* f^-1(A) is by definition a subset of the domain (it consists of elements x in A such that...)

But you were talking about Fatou and Julia sets, which are

So, let [tex]B\subseteq A[/tex].

Recall that [tex]f(B):=\{f(x)\ |\ x\in B\}[/tex]. Therefore, the statement [tex]f(B)\subseteq B[/tex] means that [tex]f(x)\in B[/tex] for all [tex]x\in B[/tex]. Hence, the statement [tex]f(B)\subseteq B[/tex] is equivalent to [tex]x\in B\Rightarrow f(x)\in B[/tex].

Recall that [tex]f^{-1}(B):=\{x\in A\ |\ f(x)\in B\}[/tex]. Therefore, the statement [tex]f^{-1}(B)\subseteq B[/tex] means that [tex]x\in B[/tex] for all [tex]f(x)\in B[/tex]. Hence, the statement [tex]f^{-1}(B)\subseteq B[/tex] is equivalent to [tex]f(x)\in B\Rightarrow x\in B[/tex].

Together: ([tex]f(B)\subseteq B[/tex] AND [tex]f^{-1}(B)\subseteq B[/tex]) is equivalent to ([tex]x\in B\Rightarrow f(x)\in B[/tex] AND [tex]f(x)\in B\Rightarrow x\in B[/tex]), and the last is equivalent to [tex]x\in B\Leftrightarrow f(x)\in B[/tex].

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Wow. That was very helpful. Thanks a lot.

* f(A) is just the image of f, and by definition the image of f is a subset of the codomain.

* f^-1(A) is by definition a subset of the domain (it consists of elements x in A such that...)

But you were talking about Fatou and Julia sets, which aresubsetsof domain and codomain.

So, let [tex]B\subseteq A[/tex].

Recall that [tex]f(B):=\{f(x)\ |\ x\in B\}[/tex]. Therefore, the statement [tex]f(B)\subseteq B[/tex] means that [tex]f(x)\in B[/tex] for all [tex]x\in B[/tex]. Hence, the statement [tex]f(B)\subseteq B[/tex] is equivalent to [tex]x\in B\Rightarrow f(x)\in B[/tex].

Recall that [tex]f^{-1}(B):=\{x\in A\ |\ f(x)\in B\}[/tex]. Therefore, the statement [tex]f^{-1}(B)\subseteq B[/tex] means that [tex]x\in B[/tex] for all [tex]f(x)\in B[/tex]. Hence, the statement [tex]f^{-1}(B)\subseteq B[/tex] is equivalent to [tex]f(x)\in B\Rightarrow x\in B[/tex].

Together: ([tex]f(B)\subseteq B[/tex] AND [tex]f^{-1}(B)\subseteq B[/tex]) is equivalent to ([tex]x\in B\Rightarrow f(x)\in B[/tex] AND [tex]f(x)\in B\Rightarrow x\in B[/tex]), and the last is equivalent to [tex]x\in B\Leftrightarrow f(x)\in B[/tex].

Here's another question, and I think if I'm right here, I can leave this behind: Does [tex]F \subseteq (f(F^c))^c[/tex] imply [tex]F \subseteq f(F)[/tex]?

- #9

Landau

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You're welcome!Wow. That was very helpful. Thanks a lot.

Too bad, this is not true. A very simple counter-example:Here's another question, and I think if I'm right here, I can leave this behind: Does [tex]F \subseteq (f(F^c))^c[/tex] imply [tex]F \subseteq f(F)[/tex]?

Take A={1,2,3}, define f:A->A by f(1)=f(2)=2 and f(3)=3. For the subset F={1,2} we now have the following:

f(F)={2}

f(F^c)^c=A\{f(3}={1,2,3}\{3}={1,2}.

Hence [tex]F\subseteq f(F^c)^c[/tex] ({1,2} is contained in {1,2}), but [tex]F\subseteq f(F)[/tex] does NOT hold ({1,2} is NOT contained in {2}).

- #10

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Yeah, I actually thought about this complication after I'd posted. But what if we assume [tex]f[/tex] is onto? Do we still have the same problem?Too bad, this is not true. A very simple counter-example:

Take A={1,2,3}, define f:A->A by f(1)=f(2)=2 and f(3)=3. For the subset F={1,2} we now have the following:

f(F)={2}

f(F^c)^c=A\{f(3}={1,2,3}\{3}={1,2}.

Hence [tex]F\subseteq f(F^c)^c[/tex] ({1,2} is contained in {1,2}), but [tex]F\subseteq f(F)[/tex] does NOT hold ({1,2} is NOT contained in {2}).

- #11

Landau

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No, then it's true. Proof:But what if we assume [tex]f[/tex] is onto? Do we still have the same problem?

Assume [tex]F\subseteq (f(F^c))^c[/tex]. This means [tex]x\in F\Rightarrow (\forall y\in F^c: x\neq f(y))[/tex].

Let [tex]x\in F[/tex]. Since f is onto, there exists [tex]z\in A[/tex] such that [tex]x=f(z)[/tex]. So [tex]z\in A\backslash F^c=F[/tex], from which it follows that [tex]x\in f(F)[/tex]. We have proven [tex]x\in F\Rightarrow x\in f(F)[/tex], i.e. [tex]F\subseteq f(F)[/tex].

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