filter

**sidi** · December 14th 2010, 12:32 PM

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

**Drexel28** · December 14th 2010, 01:10 PM

Originally Posted by sidi

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?

**sidi** · December 14th 2010, 01:27 PM

really! Thanks a lot, I guess I am remade....thanks

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