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If $A\subseteq B$ under what conditions is $f^{-1}(A) \subseteq f^{-1}(B)$, where $f^{-1}$ is the preimage, not the inverse.

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First, we should be clear about the definition of $f$. Suppose $X, Y$ are sets and $f: X \to Y$ is a map. Suppose $A, B \subseteq Y$ with $A \subseteq B$.

It's always true that $f^{-1}(A)$ is contained in $f^{-1}(B)$, and this should be clear by the definition of preimage.

$f^{-1}(A)$ is the stuff in $X$ that is mapped into $A$. But since $A \subseteq B$, if stuff in $X$ is mapped into $A$, then that same stuff is mapped into $B$ because $A$ is a subset of $B$. Then that means the stuff mapped into $A$ is a subset of the stuff mapped into $B$, i.e., $f^{-1}(A) \subseteq f^{-1}(B)$.

Bonus question for you: If $A \subseteq B$, when is $f^{-1}(A) = f^{-1}(B)$?

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    $\begingroup$ If and only if $A=B$? $\endgroup$ – usainlightning Feb 10 '15 at 19:05
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    $\begingroup$ $f^{-1}(A) =f^{-1}(B) \iff ${$x \in \mathbb{R}|f(x) \in A \cap B $}, Therefore as long as $f^{-1}(A/B)$ and $ f^{-1}(B/A)$ is the empty set this is true. $\endgroup$ – usainlightning Feb 10 '15 at 19:51
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    $\begingroup$ I didn't see your hint, am i still correct? $\endgroup$ – usainlightning Feb 10 '15 at 19:52
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    $\begingroup$ @usainlightning I'm not sure what you mean with your $\iff$. You have a statement on the left hand side, but a set on the right hand side. Here is what I was looking for as a proof: By my hint, since $B = A \cup (B - A)$, then $f^{-1}(B) = f^{-1}(A) \cup f^{-1}(B - A)$. So if $f^{-1}(B - A) = \emptyset$, then we get $f^{-1}(B) = f^{-1}(A) \cup \emptyset = f^{-1}(A)$. $\endgroup$ – layman Feb 10 '15 at 20:32
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    $\begingroup$ Yes sorry that is obvious $\endgroup$ – usainlightning Feb 10 '15 at 21:14

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