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  1. #1
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    filter

    Can anybody help me with this problem please?

    f is mapping from X onto Y, F is filter on X, f(F) need not to be filter on Y.

    I can't see no counterexample. Thank you very much
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sidi View Post
    Can anybody help me with this problem please?

    f is mapping from X onto Y, F is filter on X, f(F) need not to be filter on Y.

    I can't see no counterexample. Thank you very much
    Let \mathcal{F} be a filter on X. Clearly since X\in\mathcal{F} then Y=f(X)\in f\left(\mathcal{F}\right). Clearly \varnothing\notin f\left(\mathcal{F}\right). Now, suppose that f\left(E\right)\in f\left(\mathcal{F}\right) and let G\subseteq f\left(E\right). Then, f^{-1}(G)\subseteq E and thus f^{-1}(G)\in \mathcal{F}. Therefore, f\left(f^{-1}(G)\right)=G\in\mathcal{F}. So how many aximos are there left to violate?
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  3. #3
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    really! Thanks a lot, I guess I am remade....thanks
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