Inverse map is closed under complementation

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SUMMARY

The discussion centers on the mathematical proof of the equality f^-1(E^c) = (f^-1(E))^c, where f is a map from set X to set Y and E is a subset of Y. The proof involves demonstrating that f^-1(E^c) is a subset of (f^-1(E))^c and vice versa. The user outlines the steps to show that if x is in f^-1(E^c), then it must also be in (f^-1(E))^c, establishing the first inclusion. The discussion encourages further exploration of the second inclusion to complete the proof.

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carr1
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f^-1 (E^c) = (f^-1(E))^c where f is map from X to Y and E is in Y.
Prove equality is true.
 
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how would I show the inverse map on the left is a subset of the inverse map on the right? and vice versa?
 
To show two sets are equal we show each is contained in the other, hence we must show $f^{-1}(E^c) \subseteq (f^{-1}(E))^c$ and $(f^{-1}(E))^c \subseteq f^{-1}(E^c)$. To do this we take an element in one of them and show it is also in the other. I'm going to do the first inclusion.

Let $x \in f^{-1}(E^c)$. By the definition of inverse image we know that $f(x) \in E^c$, but this means that $f(x) \notin E$. Hence $x \notin f^{-1}(E)$ and we conclude that $x \in (f^{-1}(E))^c$. Therefore $f^{-1}(E^c) \subseteq (f^{-1}(E))^c$.

Try the second inclusion. :)

Best wishes,

Fantini.
 

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