# filter

• Dec 14th 2010, 12:32 PM
sidi
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
• Dec 14th 2010, 01:10 PM
Drexel28
Quote:

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 $\displaystyle \mathcal{F}$ 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? ;)
• Dec 14th 2010, 01:27 PM
sidi
really! Thanks a lot, I guess I am remade....thanks