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Let $\scr{F}$ be free filter ($\cap\scr{F}=\emptyset$) on a countable set $X$ and $B\in\scr{F}$. We define the trace of $\scr{F}$ on $B$ as follows $\mathscr{F}_B=\{Y\cap B:~Y\in\scr{F}\}$. $\scr{F}$ and $\mathscr{F}_B$ are isomorphic in the case of Frechet filter or in the case of any ultrafilter. Are $\scr{F}$ and $\mathscr{F}_B$ isomorphic in all cases ?

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The answer is negative. You can get any infinite $B$ with infinite complement. Then let $\scr{N}$ be the Frechet filter on $B$ and $\scr{F}$ be the filter on $X$ generated by $\scr{N}$. Then $\mathscr{F}_B = \scr{N}$ and $\mathscr{F}\not\simeq \mathscr{F}_B$ because for any bijection $\phi:X\to B$ the set $\phi(B)$ has infinite complement and thus $\phi(B)\notin\mathscr{F}_B$.

But if $B$ has infinite complement in $X$ and has subset $B\supset B_0\in\scr{F}$ such that $B-B_0$ is infinite, then $\mathscr{F}\simeq \mathscr{F}_B$.

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