# Confirming proofs of properties of preimages

I do not think that I made terrible mistakes, nevertheless a conforming word would be good for me. Thank you!

Let $f\colon X\to Y$ be a map and $(B_i)_{i\in I}$ a family of subsets of $Y$ ($I$ is any index set). Overmore consider any $A,B\subset Y$. Show: \begin{align} &(1)~f^{-1}(\bigcup_{i\in I}B_i)=\bigcup_{i\in I}f^{-1}(B_i)\\ &(2)~f^{-1}(\bigcap_{i\in I} B_i)=\bigcap_{i\in I}f^{-1}(B_i)\\ &(3)~(f^{-1}(A))^C=f^{-1}(A^C)\\ &(4)~f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B)\\ &(5)~f^{-1}(A\Delta B)=f^{-1}(A)\Delta f^{-1}(B) \end{align}

My proofs:

(1) "$\subset$":

\begin{align} x\in f^{-1}(\bigcup_{i\in I}B_i) &\Rightarrow\exists~j\in I: f(x)\in B_j\\&\Rightarrow x\in f^{-1}(B_j)\\&\Rightarrow x\in\bigcup_{i\in I}f^{-1}(B_i) \end{align} "$\supset$": \begin{align} x\in\bigcup_{i\in I}f^{-1}(B_i) &\Rightarrow\exists j\in I: x\in f^{-1}(B_j)\\&\Rightarrow f(x)\in B_j\\&\Rightarrow f(x)\in\bigcup_{i\in I}B_i\\&\Rightarrow x\in f^{-1}(\bigcup_{i\in I}B_i)~~~\Box\end{align} (2) "$\subset$": \begin{align} x\in f^{-1}(\bigcap_{i\in I} &\Rightarrow f(x)\in\bigcap_{i\in I}B_i\\&\Rightarrow f(x)\in B_i~\forall~i\in I\\&\Rightarrow x\in f^{-1}(B)~\forall~i\in I\\&\Rightarrow x\in\bigcap_{i\in I}f^{-1}(B_i) \end{align} "$\supset$": \begin{align} x\in\bigcap_{i\in I}f^{-1}(B_i) &\Rightarrow x\in f^{-1}(B_i)~\forall~i\in I\\&\Rightarrow f(x)\in B_i~\forall~i\in I\\&\Rightarrow f(x)\in\bigcap_{i\in I}B_i\\&\Rightarrow x\in f^{-1}(\bigcap_{i\in I}B_i)~~~\Box \end{align} (3) "$\subset$": \begin{align} x\in (f^{-1}(A))^C=X\setminus f^{-1}(A) &\Rightarrow x\in X\wedge x\notin f^{-1}(A)\\&\Rightarrow x\in X\wedge f(x)\notin A\\&\Rightarrow x\in X\wedge T(x)\in A^C\\&\Rightarrow x\in X\wedge x\in f^{-1}(A^C)\\&\Rightarrow x\in X\cap f^{-1}(A^C)=f^{-1}(A^C) \end{align} "$\supset$": \begin{align} x\in f^{-1}(A^C)=Y\ &\Rightarrow f(x)\in A^C=Y\setminus A\\&\Rightarrow f(x)\in Y\wedge f(x)\notin A\\&\Rightarrow x\in f^{-1}(Y)=X\wedge x\notin f^{-1}(A)\\&\Rightarrow x\in X\setminus f^{-1}(A)=(f^{-1}(A))^C~~~\Box \end{align} (4) "$\subset$": \begin{align} x\in f^{-1}(A\setminus B) &\Rightarrow f(x)\in A\wedge f(x)\notin B\\&\Rightarrow x\in f^{-1}(A)\wedge x\notin f^{-1}(B)\\&\Rightarrow x\in f^{-1}(A)\setminus f^{-1}(B) \end{align} "$\supset$": \begin{align} x\in f^{-1}(A)\setminus f^{-1}(B) &\Rightarrow x\in f^{-1}(A)\wedge f(x)\notin f^{-1}(B)\\&\Rightarrow f(x)\in A\wedge f(x)\notin B\\&\Rightarrow f(x)\in A\setminus B\\&\Rightarrow x\in f^{-1}(A\setminus B)~~~\Box \end{align} (5) "$\subset$": \begin{align} x\in f^{-1}(A\Delta B) &\Rightarrow f(x)\in (A\setminus B)\cup (B\setminus A)\\&\Rightarrow f(x)\in (A\setminus B)\vee f(x)\in B\setminus A \end{align} I. $f(x)\in A\setminus B$: \begin{align} &\Rightarrow f(x)\in A\wedge f(x)\notin B\\&\Rightarrow x\in f^{-1}(A)\wedge x\notin f^{-1}(B)\\&\Rightarrow x\in f^{-1}(A)\setminus f^{-1}(B)\\&\Rightarrow x\in (f^{-1}(A)\setminus f^{-1}(B))\cup (f^{-1}(B)\setminus f^{-1}(A))=f^{-1}(A)\Delta f^{-1}(B) \end{align} II. $f(x)\in B\setminus A$: \begin{align} &\Rightarrow f(x)\in B\wedge f(x)\notin A\\&\Rightarrow x\in f^{-1}(B)\setminus f^{-1}(A)\\&\Rightarrow x\in f^{-1}(A)\Delta f^{-1}(B) \end{align} "$\supset$": \begin{align} x\in f^{-1}(A)\Delta f^{-1}(B) &\Rightarrow x\in (f^{-1}(A)\setminus f^{-1}(B))\vee (f^{-1}(B)\setminus f^{-1}(A)) \end{align} I. $x\in f^{-1}(A)\setminus f^{-1}(B)$: \begin{align} &\Rightarrow x\in f^{-1}(A)\wedge x\notin f^{-1}(B)\\&\Rightarrow f(x)\in A\wedge f(x)\notin B\\&\Rightarrow f(x)\in A\setminus B\\&\Rightarrow x\in f^{-1}(A\setminus B)\cup f^{-1}(B\setminus A)\overset{\text{with}}{\underset{\text{(1)}}{=}}f^{-1}((A\setminus B)\cup (B\setminus A))=f^{-1}(A\Delta B) \end{align} II. $x\in f^{-1}(B)\setminus f^{-1}(A)$: \begin{align} &\Rightarrow x\in f^{-1}(B)\wedge x\notin f^{-1}(A)\\&\Rightarrow f(x)\in B\wedge f(x)\notin A\\&\Rightarrow f(x)\in B\setminus A\\&\Rightarrow x\in f^{-1}(B\setminus A)\\&\Rightarrow x\in f^{-1}(B\setminus A)\cup f^{-1}(A\setminus B)\overset{\text{with}}{\underset{\text{(1)}}{=}}f^{-1}((B\setminus A)\cup (A\setminus B))=f^{-1}(A\Delta B)~~~\Box \end{align}

• You can find links to some other posts about (som of) these identities here. – Martin Sleziak Jan 18 '16 at 8:42

Nothing blatantly wrong that I can see (apart from a few typos, like missing "$B_i)$" in the $\subset$ direction of part 2, "$T(x)$" in the $\subset$ direction of part 3, and a random "$=Y$" in the $\supset$ direction of part 3), but you're making it a bit too complicated in your proofs of the last two. Indeed, $$f^{-1}(A\setminus B)=f^{-1}(A\cap B^c)\overset{(2)}{=}f^{-1}(A)\cap f^{-1}(B^c)\overset{(3)}{=}f^{-1}(A)\cap(f^{-1}(B))^c=f^{-1}(A)\setminus f^{-1}(B),$$ and \begin{align}f^{-1}(A\triangle B) &= f^{-1}\bigl((A\setminus B)\cup(B\setminus A)\bigr)\\ &\overset{(1)}{=}f^{-1}(A\setminus B)\cup f^{-1}(B\setminus A)\\ &\overset{(4)}{=}\bigl(f^{-1}(A)\setminus f^{-1}(B)\bigr)\cup\bigl(f^{-1}(B)\setminus f^{-1}(A)\bigr)\\ &= f^{-1}(A)\triangle f^{-1}(B),\end{align} with no need for element-chasing, double-inclusion, or breaking into cases.
Also, you could do parts 1 through 3 as $\iff$ all the way, rather than a double-inclusion proof.