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Prove that intersection of principal filters on $I$ is still a principal filter.

Let S be a nonempty set and C be a nonempty subset. Then {C} is a filter base. The filter it generates (i.e., the collection of all subsets containing C) is called the principal filter generated by C. Sorry for posting in this way, I understand it's not polite, but could you give any tip-off.

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    $\begingroup$ What is $I$? What is the definition of a principal filter (on $I$)? What have you tried? $\endgroup$ – Cameron Buie May 29 '13 at 14:56
  • $\begingroup$ I is an arbitrary set. Actually I was asked to post it here for any ideas. $\endgroup$ – kushtibargo May 29 '13 at 14:58
  • $\begingroup$ That answers one of my questions. What about the other two? It isn't a problem to post such things here, and people will be glad to help you, but we'll be able to give you a better answer if we know more about what you've done, what your definitions are, etc. $\endgroup$ – Cameron Buie May 29 '13 at 15:00
  • $\begingroup$ Let S be a nonempty set and C be a nonempty subset. Then {C} is a filter base. The filter it generates (i.e., the collection of all subsets containing C) is called the principal filter generated by C. Sorry for posting in this way, I understand it's not polite, but could you give any tip-off. $\endgroup$ – kushtibargo May 29 '13 at 15:06
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    $\begingroup$ @Minimus: Using principle is widespread e.g. for filters and bundles, but it's simply wrong. $\endgroup$ – Martin May 29 '13 at 21:09
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Some preliminary definitions:

“D is a filter” = df.
i.D $\subseteq$S(I) for some nonempty set I, where S(I) is the powerset of I, &
ii. I$\in$D;
iii. Given X, Y$\in$D, then X$\cap$Y $\in$D;
iv. Given X $\in$ D & X$\subseteq$Z $\subseteq$I then Z$\subseteq$ D.

“D is principal” =df. D is a filter and D = {X$\subseteq$I:Y $\subseteq$ X}.

  1. Prove: D is principal iff $\cap$D$\in$D.

Left to Right of “iff”: Suppose that D is a principal filter. Then D = {X$\subseteq$I: Y$\subseteq$ X}. So take $\cap$D. Clearly $\cap$D = Y, and Y $\in$ D (Y$\subseteq$ Y, and Y is thus one of the subsets of I having Y as subset). Thus $\cap$D$\in$D. <END OF LEFT TO RIGHT>.

Right to Left of “iff”: Suppose that $\cap$D$\in$D, and that D is a filter. We show D is principal. Thus by assumption D $\subseteq$ S(I), and so $\cap$ D$\in$S(I) hence $\cap$ D$ $ $\subset$ I. Thus $\exists$Y: D$\subseteq$Y & Y$\subseteq$I. Now call {Y: D $\subseteq$ Y & Y $\subseteq$ I} = Z. Clearly $\cap$D $\in$ Z. But $\cap$D $\in$ D as well (by initial assumption); so by extensionality D = Z; and so we have the appropriate D, so that D must be principal. <END OF RIGHT TO LEFT>.

<END OF PROOF>.

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    $\begingroup$ Please write the content of your answer here, not on any external website. Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. $\endgroup$ – Zev Chonoles Jun 10 '13 at 20:32
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Take any two non-empty subsets $A,B$ of $I$, and let $\mathcal E,\mathcal F$ the principal filters on $I$ that they generate. Then $\mathcal E\cap\mathcal F$ is the set of subsets $C$ of $I$ that contain both $A$ and $B$ (that is, having both $A$ and $B$ as subsets). If $C\subseteq I$ is such that $A,B\subseteq C,$ what can you say with regard to $A\cup B$ and $C$? If $A\cup B\subseteq C,$ what can you say with regard to $A,B$ and $C$?

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  • $\begingroup$ M={A, B}, so M generates filter with base A and B? $\endgroup$ – kushtibargo May 29 '13 at 15:23
  • $\begingroup$ That has $A$ and $B$ as elements. When we say that a set $S$ "contains" a set $T$, we mean that $T\subseteq S.$ $\endgroup$ – Cameron Buie May 29 '13 at 15:57
  • $\begingroup$ @kushtibargo: Suppose that $A\subseteq S$ and $B\subseteq S$; what can you say about $A\cup B$? $\endgroup$ – Brian M. Scott May 29 '13 at 19:04

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