# Mergers and ultrafilters

F - filter over I. Please, prove, F ultrafilter <=> when $\forall$ X, Y $\subseteq$ I, if X $\notin$ F, Y $\notin$ F, $\Rightarrow$ X $\cup$ Y $\notin$ F.

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@MariyaSmit: Welcome to MSE! It also helps readability to format questions using MathJax (see FAQ). Regards – Amzoti May 27 '13 at 13:35
How do you define an ultrafilter? – Asaf Karagila May 27 '13 at 13:38
F - ultrafilter over I, if $\forall$ A $\subseteq$ I, A $\in$ F or I\A $\in$ F – Flam May 27 '13 at 13:44
Suppose first that $\mathscr{F}$ is an ultrafilter on $I$, that $X,Y\subseteq I$, and that $X,Y\notin\mathscr{F}$. Then $I\setminus X,I\setminus Y\in\mathscr{F}$, so $$I\setminus(X\cup Y)=(I\setminus X)\cap(I\setminus Y)\in\mathscr{F}\;,$$ and therefore $$X\cup Y=I\setminus(I\setminus(X\cup Y))\notin\mathscr{F}\;.$$ Conversely, if $\mathscr{F}$ is not an ultrafilter, then there is some $X\subseteq I$ such that $X\notin\mathscr{F}$ and $I\setminus X\notin\mathscr{F}$. But then $X\cup(I\setminus X)=I$, and ... ? – Brian M. Scott May 27 '13 at 19:27

Hint: If $\cal F$ is an ultrafilter, pick $X,Y\notin\cal F$ and use the fact that their complements are in $\cal F$ and it is closed under intersections; in the other direction, if $\cal F$ is not an ultrafilter there is some $X$ such that $X,I\setminus X\notin\cal F$. Use these two sets to contradict the statement you have on the RHS.