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Thread: filter

  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 $\displaystyle \mathcal{F}$ be a filter on $\displaystyle X$. Clearly since $\displaystyle X\in\mathcal{F}$ then $\displaystyle Y=f(X)\in f\left(\mathcal{F}\right)$. Clearly $\displaystyle \varnothing\notin f\left(\mathcal{F}\right)$. Now, suppose that $\displaystyle f\left(E\right)\in f\left(\mathcal{F}\right)$ and let $\displaystyle G\subseteq f\left(E\right)$. Then, $\displaystyle f^{-1}(G)\subseteq E$ and thus $\displaystyle f^{-1}(G)\in \mathcal{F}$. Therefore, $\displaystyle f\left(f^{-1}(G)\right)=G\in\mathcal{F}$. 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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