MHB Inverse map is closed under complementation

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The discussion centers on proving the equality f^-1(E^c) = (f^-1(E))^c, where f is a map from X to Y and E is a subset of Y. Participants emphasize the need to show that each set is contained within the other to establish equality. The first inclusion is demonstrated by taking an element from f^-1(E^c) and showing it belongs to (f^-1(E))^c. The next step involves proving the reverse inclusion, starting with an element from (f^-1(E))^c. The conversation highlights the logical approach to proving set inclusions in the context of inverse images.
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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.
 

Thread 'About the definition of topological manifold using closed sets'

We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.
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