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Let $f:S\to T$ be a function. Show that for each subset $B \subseteq T$, we have $f[f^{-1}[B]] = B$ iff $B \subseteq \operatorname{Rng}(f)$.

here is what I have :

To show that $f^{-1}[f[B]] =B\Leftrightarrow B\subseteq \operatorname{Rng}(f)$ : \begin{align}f^{-1}[f[B]] = B & \Leftrightarrow f(b_1)=f(b_2)\qquad \text{for some } b_1,b_2\in B \\ & \Leftrightarrow b_1=b_2 \\ & \Leftrightarrow f(b)\in B \\ & \Leftrightarrow\operatorname{Rng}(f)=B \\ & \Leftrightarrow B\subseteq \operatorname{Rng}(f) \end{align}

Arnaud D.
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user5428151
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  • A proof should contain justifications of the steps taken. This is also what you should do: can you justify each step? Also, you keep writing 'iff', but are you sure all of these are really 'iff'? (Example: you write $f(b_1) =f(b_2)$ iff $b_1 = b_2$, but only one direction is valid in general, the other one requires the function to be injective in general...) – Student Mar 28 '18 at 10:19
  • Note that $B\subset T$, so if $b\in B$, it doesn't make sense to speak of $f(b)$. – Arnaud D. Mar 28 '18 at 10:20
  • See also this question and those that are linked to it : https://math.stackexchange.com/questions/359693/overview-of-basic-results-about-images-and-preimages – Arnaud D. Mar 28 '18 at 10:21
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    @ArnaudD. OP asks for possible mistakes in his proof. "$f^{-1}[f[B]]$" at the beginning is part of his proof, so nobody should correct this in an edit but OP him/herself. – GNUSupporter 8964民主女神 地下教會 Mar 28 '18 at 10:47
  • @GNUSupporter I hadn't noticed the difference. I suspect that it's only a typo, but in doubt I've changed it to match the original. – Arnaud D. Mar 28 '18 at 10:58
  • @ArnaudD. Thanks for your edit. – GNUSupporter 8964民主女神 地下教會 Mar 28 '18 at 11:00

3 Answers3

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Your proof has some problems. As I commented on your question, I think you can find this out yourself by trying to justify each step you take. This is what you need to do in a proof: write down exactly why you can make some steps.

In general, if we have to prove an 'iff' statement, this is done by proving to implications ($\Rightarrow$). In your case, this means that we need to prove two things:

  1. $B \subset \operatorname{Rng}(f) \Rightarrow f[f^{-1}(B)] = B$ and
  2. $f[f^{-1}(B)] = B \Rightarrow B \subset \operatorname{Rng}(f)$.

The first part asks you to prove an equality of sets. This is usually done by showing that the left hand side is a subset of the right hand side and the right hand side is a subset of the left hand side: $f[f^{-1}(B)] \subset B$ and $B \subset f[f^{-1}(B)]$. (this is also called 'proving two inclusions'). Inclusions are generally proved by taking an arbitrary element from the first set and showing it is also an element of the second set.

For the prove itself:

  1. assume that $B \subset \operatorname{Rng}(f)$. We need to prove two inclusions.

Let $t \in f[f^{-1}(B)]$, then there is an element $s \in f^{-1}(B)$ such that $f(s) = t$. By definition of $s$, we have that $t = f(s) \in B$ and therefore $f[f^{-1}(B)] \subset B$.

Let $t \in B$. Since $B \subset \operatorname{Rng}(f)$, there is an $s \in S$ such that $t = f(s)$. Since $f(s) \in B$, $s \in f^{-1}(B)$ and we have $t = f(s) \in f[f^{-1}(B)]$. This proves that $B \subset f[f^{-1}(B)]$ and concludes the prove of this implication.

  1. Assume that $f[f^{-1}(B)] = B$. We need to prove that $B \subset \operatorname{Rng}(f)$.

I will let you do this one. If you have any problems, let me know.

Student
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1

(This addresses the proof in the picture in the original version of the question.)

You've miswritten $f[f^{-1}[B]]$ as $f^{-1}[f[B]]$ in the beginning of the proof. Since $B \subseteq T$ and $T$ and the codomain of $f$, you can't write $f[B]$ in general.

You proof lost me.

\begin{align} f[f^{-1}[B]] = B & \iff \dots \\ & \iff \forall\, b \in B, \exists\, s \in S \text{ s.t. } f(s) = b \\ & \iff B\subseteq \operatorname{Rng}(f) \end{align}

  • If part: Suppose $f[f^{-1}[B]] = B$. Let $b \in B = f[f^{-1}[B]]$. There exists $s \in f^{-1}[B]$ such that $f(s) = b$. This proves $B\subseteq \operatorname{Rng}(f)$.

  • Only-if part: Suppose $B\subseteq \operatorname{Rng}(f)$.

    • $f[f^{-1}[B]] \subseteq B$: This is tautology. $$f^{-1}[B] = \{s \in S \mid f(s) \in B\} \tag{*} \label{*}$$ Applying $f$ on $f^{-1}[B]$, you map the members by $f$, but the predicate $f(s) \in B$ is unchanged.
    • $B \subseteq f[f^{-1}[B]]$: Let $b \in B$. By assumption $B\subseteq \operatorname{Rng}(f)$, there exists $s \in S$ such that $f(s) = b \in B$. By the very definition of $f^{-1}[B]$ in \eqref{*}, $s \in f^{-1}[B]$, so $b = f(s) \in f[f^{-1}[B]]$. Since the choice of $b \in B$ is arbitrary, this concludes the proof.
GNUSupporter 8964民主女神 地下教會
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Let f' = $f^{-1}.$ ff'(A) = f(dom f) $\cap$ f(A).
Proof. y in ff'(A)
iff some x in f'(A) with y = f(x)
iff some x with f(x) in f(A) and y = f(x)
iff y in f(dom f) and y in f(A).

From this equation your problem can easily be proved.

William Elliot
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