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Basic (I think) image/preimage questions

  1. Mar 13, 2010 #1
    Suppose I know that [tex]x\in A \Leftrightarrow f(x)\in A[/tex]. Can someone explain why I know, on the strength of this, that [tex]f(A) \subseteq A[/tex] and [tex]f^{-1}(A) \subseteq A[/tex]?
     
  2. jcsd
  3. Mar 13, 2010 #2
    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]
     
    Last edited: Mar 14, 2010
  4. Mar 14, 2010 #3

    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...?
     
  5. Mar 14, 2010 #4
    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].

    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 Complex Analysis."
     
  6. Mar 14, 2010 #5

    Landau

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    I didn't mean for you to explain the specific problem you're working on (Fatou, Julia sets, rational functions, etc.), but the general A and f.

    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. Mar 14, 2010 #6
    Ok...well, [tex]A = \mathbb{C}^*[/tex], and [tex]f: \mathbb C^* \to \mathbb C^*[/tex] is a rational function. Does that help?
     
  8. Mar 14, 2010 #7

    Landau

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    For a function [tex]f:A\to A[/tex], the statements [tex]f(A)\subseteq A[/tex] and [tex]f^{-1}(A)\subseteq A[/tex] are trivial:
    * 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 subsets of 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].
     
  9. Mar 15, 2010 #8
    Wow. That was very helpful. Thanks a lot.

    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]?
     
  10. Mar 15, 2010 #9

    Landau

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    You're welcome!
    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. Mar 15, 2010 #10
    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?
     
  12. Mar 15, 2010 #11

    Landau

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    No, then it's true. Proof:

    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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