MHB Inverse map is closed under complementation

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.
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...

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