Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  draks ... May 27 '13 at 13:27
    
@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 –  Mariya Smit 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
add comment

1 Answer 1

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.

share|improve this answer
    
Thanks for answer! –  Mariya Smit May 27 '13 at 13:50
    
1)How i can make conversion by I\X ∈ F and I\Y ∈ F to I(X ∪ Y) ∈ F. 2)And sorry, can you explain me "in the other direction" a little more detail? –  Mariya Smit May 27 '13 at 14:00
    
Use DeMorgan's laws. –  Asaf Karagila May 27 '13 at 14:04
    
Thanks for answer! –  Mariya Smit May 27 '13 at 14:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.