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$f(f^{-1}(V))\subset V$

Proof:"To see $f(f^{-1}(V)) \subset V$, let $x \in f^{-1}(V)$. Then by definition, $f(x) \in V$. In other words, for all $x \in f^{-1}(V), f(x) \in V$. This means that $$\{f(x) : x \in f^{-1}(V)\} =: f(f^{-1}(V)) \subset V$$ where the equality is the definition of $f(f^{-1}(V))$". By Marcus M.

I have tried to find examples of function in which this property holds. I thought that any arbitrary function would suffice.

If I take $f(x)=x^2$, and I choose $A=(0,4)$, $f(f^{-1}((0,4)))= f((0,2))=(0,4)$ which contradicts the claim.

Question:

What am I doing wrong? Can someone provide me example where this property holds? I am looking for intuition on this issue.

Thanks in advance!

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  • $\begingroup$ Why do you think your example contradicts the claim? You have $(0,4)\subset (0,4)$, so the claim holds. $\endgroup$ – Arnaud D. Oct 1 '17 at 10:42
  • $\begingroup$ Note that $f^{-1}((0,4))=(-2,0)\cup(0,2)$. Still, it doesn't contradict the claim. $\endgroup$ – Arthur Oct 1 '17 at 10:44
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First, you cannot just say $f(x) = x^2$, especially when you work with inverse images. You must say on which set your function is defined. Is it $\mathbb{R}$ ? $\mathbb{R}_{\ge 0}$ ? $\mathbb{R}_{>0}$ ? $\mathbb{C}$.

So that your reasoning is correct, let us assume we take $f : \mathbb{R}_{>0} \to \mathbb{R}, x \mapsto x^2$ (because otherwise, you forgot the negative values : for example $-1$ is mapped to $1$). Even then, there is no contradiction : $(0,4)$ is a subset of $(0,4)$. The expression $A\subset B$ does not imply that $A$ is a strict subset of $B$. For example, $A \subset A$ is always true.

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  • $\begingroup$ Are there functions in which the counter-dominion of pre-image gives an interval smaller the interval considered? $\endgroup$ – Pedro Gomes Oct 1 '17 at 11:30
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    $\begingroup$ Yes, the idea being $V$ is not in the image of $f$. Taking this time $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$, we have $f(f{-1}([-1,1])) = [0,1]$ which is a strict subset of $[-1,1]$. $\endgroup$ – Junkyards Oct 1 '17 at 14:23
  • $\begingroup$ @PedroGomes One can show that the equality $f(f^{-1}(A))=A$ holds if and only if $A\subset \operatorname{im}(f)$. So if $A$ is not included in the image of $f$ the inclusion will be strict. $\endgroup$ – Arnaud D. Oct 1 '17 at 14:51

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