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Prove $f^{-1}(\overline{A}∩B)⊆f^{-1}(\overline{A})∩f^{-1}(B)$
Where f: P→Q and A and B are non empty subsets of Q

So far I have:
$f^{-1}(\overline{A}∩B)⊆f^{-1}(\overline{A})∩f^{-1}(B)$
$f^{-1}(\overline{A}∩B)↔{x∈f^{-1}(\overline{A})∧x∈f^{-1}(B)}$
$f^{-1}(\overline{A}∩B)↔{x∈f^{-1}(\overline{A})}∧{x∈f^{-1}(B)}$
$f^{-1}(\overline{A}∩B)↔f^{-1}(\overline{A})∩f^{-1}(B)$
$f^{-1}(\overline{A}∩B)⊆f^{-1}(\overline{A})∩f^{-1}(B)$

but I'm unsure if this is the correct way to prove it
Any help would be appreciated.

June Tejada
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    Note that the theorem holds for any sets and you need not worry about the closure of the set – FShrike May 11 '22 at 19:49
  • What is the point of linking to a picture of the formula you are correctly typesetting? – Arturo Magidin May 11 '22 at 19:49
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    I'm very confused by your proof attempt: it seems to both begin and end with your goal, and it has implications where one or both sides of the connective are sets instead of propositions. Sorry but to me it seems completely incomprehensible – Stephen Donovan May 11 '22 at 19:53

2 Answers2

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Using the comment by @FShrike, suppose that $x \in f^{-1}( A \cap B )$. For some $y \in A \cap B$ we have $f(x) = y$. Notice that $y \in A$ and $y \in B$. Thus $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$. That is to say $x \in f^{-1}(A) \cap f^{-1}(B)$.

Jay
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You even have an equality since: \begin{align*} x\in f^{-1}(\overline{A}\cap B) &\Leftrightarrow f(x)\in\overline{A}\cap B \Leftrightarrow f(x)\in\overline{A}\land f(x)\in B \\ &\Leftrightarrow x\in f^{-1}(\overline{A})\land f^{-1}(x)\in B \Leftrightarrow x\in f^{-1}(\overline{A})\cap B. \end{align*} Other answers can be found here.

Samuel Adrian Antz
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