Basic (I think) image/preimage questions

[AxiomOfChoice]

Suppose I know that x\in A \Leftrightarrow f(x)\in A. Can someone explain why I know, on the strength of this, that f(A) \subseteq A and f^{-1}(A) \subseteq A?

 

[AxiomOfChoice]

Actually, I think I've got it (somebody please verify):

<br /> x\in f(A) \Rightarrow f^{-1}(x) \in A \Rightarrow f(f^{-1}(x)) \in A \Rightarrow x\in A<br />

<br /> x\in f^{-1}(A) \Rightarrow f(x) \in A \Rightarrow x\in A<br />

 

[Landau]

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

 

[AxiomOfChoice]

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

This is all geared toward showing that the Fatou set F and Julia set J of a rational function are completely invariant. Apparently, since F = J^c, showing that f(F) \subseteq F and f(J) \subseteq J amounts to showing f(F) \subseteq F and f^{-1}(F) \subseteq F, which is apparently equivalent to showing z_0 \in F iff f(z_0) \in F.

The above questions about general A and general f 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."

 

[Landau]

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 f(A)\subseteq A 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?

 

[AxiomOfChoice]

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 f(A)\subseteq A 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?

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

 

[Landau]

For a function f:A\to A, the statements f(A)\subseteq A and f^{-1}(A)\subseteq A 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 B\subseteq A.

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

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

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

 

[AxiomOfChoice]

For a function f:A\to A, the statements f(A)\subseteq A and f^{-1}(A)\subseteq A 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 B\subseteq A.

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

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

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

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 F \subseteq (f(F^c))^c imply F \subseteq f(F)?

 

[Landau]

Wow. That was very helpful. Thanks a lot.

You're welcome!

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

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

 

[AxiomOfChoice]

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

Yeah, I actually thought about this complication after I'd posted. But what if we assume f is onto? Do we still have the same problem?

 

[Landau]

But what if we assume f is onto? Do we still have the same problem?

No, then it's true. Proof:

Assume F\subseteq (f(F^c))^c. This means x\in F\Rightarrow (\forall y\in F^c: x\neq f(y)).
Let x\in F. Since f is onto, there exists z\in A such that x=f(z). So z\in A\backslash F^c=F, from which it follows that x\in f(F). We have proven x\in F\Rightarrow x\in f(F), i.e. F\subseteq f(F).