Let $A$ and $B$ be sets, let $X \subseteq A$ and $Y\subseteq B$ be subsets and let $f: A \rightarrow B$ be a function.

I tried to make a proof of the first ones but i don't know how to proceed for 3 and 4.

  1. Prove that $X\subseteq f^{-1}(f(X))$

If $a\in X$. Then $f(a)\in f(X)$ and by definition $a \in f^{-1}(f(x))$

  1. Prove that $f(f^{-1}(Y)) \subseteq Y$

Let $y\in f(f^{-1}(Y))$, which means that for some $x\in f^{-1}(Y) $ we have $f(x)=y$ But $x \in f^{-1}(Y)$ means that $f(y)\in Y$ by definition, so $y \in Y$.

  1. Prove that $X=f^{-1}(f(X))$ if and only if $X=f^{-1}(Z)$ for some $Z\subseteq B$
  2. Prove that $f(f^{-1}(f(X)))=f(X)$


=> X=f$^{-1}$(f(X))

Can take Z=f(X) ={b$\in $ B : b=f(a) for some a$\in$ A}


Claim: X=f$^{-1}$(Z) for some Z $\subseteq$ B implies f(X) = Z

proof: Take any element y in f(X).Hence there exists x s.t f(x)=y.

Hence $\exists$ z s.t. x=f$^{-1}$(z) implies f(x)=z,z$\in$ Z => y=z$\in$Z

similarly take z in Z.there exists x s.t. x=f$^{-1}$(z)=>f(x)=y=z

hence combining both f(X)= Z

Now, Given X = $f^{-1}(Z)$ => X = $f^{-1}(f(X))$


f(X) $\subseteq$ B

Consider the function $f^{-1}$ : B -> A

Want to apply 3 $\textit{where X is replaced by f(X), f is replaced by f$^{-1}$ and f$^{-1}$ is replaced by f}$

Hence enough to show f(X) = f(Z) for some Z $\subseteq$ A

Can Take Z = X, X $\subseteq$ A


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