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):

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

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

 

[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. $$F\subseteq f(F)$$.