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In a course in logic and set theory, we studied the concept of a Filter. We defined a filter $F \in P(S)$ on $S$ an equivalent of the following definition from Jech's Introduction to Set Theory:

(a) $S \in F$ and $\emptyset \notin F.$

(b) If $X\in F$ and $Y \in F$ then $X \cap Y \in F$.

(c) If $X \in F$ and $X \subseteq Y \subseteq S$, then $Y \in F$.

I am having trouble grasping this concept and it's meaning.

My question is, what is the intuition behind this definition, and why are these kinds of sets called filters?

Thanks

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3 Answers 3

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When you put a filter in your sink, the idea is that you filter out the big chunks of food, and let the water and the smaller chunks (which can—in principle—be washed through the pipes) go through.

You filter out the larger parts.

A filter filters out the larger sets. It is a way to say "these sets are 'large'", and the sort of make sense. The set of "everything" is definitely large, and "nothing" is definitely not; if something is larger than a large set, then it is also large; and two large sets intersect on a large set.

From a mathematical point of view you can think about this as being co-finite, or being of measure $1$ on the unit interval, or having a dense open subset (again on the unit interval). These are examples of ways where a set can be thought of as "almost everything". And that is the idea behind a filter.

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  • $\begingroup$ Since I don't read French, I can't quite read the original paper which introduced them. But I still like this analogy. $\endgroup$
    – Asaf Karagila
    Jun 12, 2018 at 0:01
  • $\begingroup$ Isn't the non-empty intersection property a bit counter-intuitive, and actually pretty amazing? It just occurred to me, maybe it's like two large people trying to get through the same door ("a large set") at the same time? $\endgroup$
    – user12802
    Jun 13, 2018 at 17:48
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    $\begingroup$ Think about "large" as being "almost everyone". If almost every store sells beer, and almost every store sells bacon, wouldn't it be weird if a lot of stores only sell one of them? $\endgroup$
    – Asaf Karagila
    Jun 13, 2018 at 18:18
  • $\begingroup$ @AsafKaragila I don's see the sink analogy in the French paper, though. Did you? $\endgroup$ Jun 15, 2018 at 13:59
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    $\begingroup$ @Henno: As I said. I can't read French. It might have been a Britta filter analogy. $\endgroup$
    – Asaf Karagila
    Jun 15, 2018 at 14:00
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It's similar to the concept of "almost everywhere". Suppose to every subset $T\subseteq S,$ you write $$ \mu(T) \begin{cases} =1 & \text{if } T\in F, \\ = 0 & \text{if } S\smallsetminus T\in F, \\ \text{is undefined} & \text{otherwise.} \end{cases} $$

Then, according to the definition of "filter", you have \begin{align} & \mu(\varnothing) = 0 \\[6pt] & \mu(S) = 1 \\[6pt] & \text{If } \mu(T_1), \mu(T_2) \text{ both exist, and } T_1\cap T_2 = \varnothing, \\ & \text{then } \mu(T_1\cup T_2) = \mu(T_1) + \mu(T_2). \end{align}

Saying $\{x\in S: P(x)\} \in F$ is the same as saying $P(x)$ for almost all $x\in S.$

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Let $X={1,2,3}$ Choose some element from $X$ say $F={{1},{1,2},{1,3},{1,2,3}}$ Then every intersection of an element of $F$ with another element in $F$ is in $F$ again. Examples ${1}∩{1,2,3}={1}$ ${1,2}∩{1,2,3}={1,2}$ ${1,3}∩{1,2,3}={1,3}$ ${1,2,3}∩{1,2,3}={1,2,3}$

Also the original $X={1,2,3}$ is also in $F$. Here $F={{1},{1,2},{1,3},{1,2,3}}$ is called the filter on $X={1,2,3}$.

Suppose we have the collection $G={{1},{1,2},{1,3},{2,3},{1,2,3}}$ Then we have ${1,3}∩{2,3}={3}$ but ${3}$ isn't in $G$. So this $G$ is not called a filter.

Now with $F={{1},{1,2},{1,3},{1,2,3}}$ can we put as any other element in it so that after placing the extra element it is still a filter? Probably not in this case. So on $X={1,2,3}$, $F={{1},{1,2},{1,3},{1,2,3}}$ is an Ultrafilter.

If we have started say with $H={{1},{1,2},{1,2,3}}$ this is still a filter on $X={1,2,3}$ but we can still add ${1,3}$ and it will still be classified as filter. So on $X={1,2,3}$

$F={{1},{1,2},{1,3},{1,2,3}}$ is an Ultrafilter but $H={{1},{1,2},{1,2,3}}$ is a filter but not an Ultrafilter.

Now suppose we have $X={1,2,3,4}$ Let $F={{1,4},{1,2,4},{1,3,4},{1,2,3,4}}$ Every in intersection of element of $F$ is in F again. We have as examples ${1,4}∩{1,4}={1,4} {1,4}∩{1,2,4}={1,4} {1,4}∩{1,3,4}={1,4} {1,2,4}∩{1,2,4}={1,2,4} {1,2,4}∩{1,3,4}={1,4} {1,3,4}∩{1,3,4}={1,3,4} {1,2,3,4}∩{1,2,3,4}={1,2,3,4}$ Also $X={1,2,3,4}$ is also in $F={{1,4},{1,2,4},{1,3,4},{1,2,3,4}}$ and the null element $∅= {}$ is not in $F$. We call $F$ a filter but not an Ultrafilter on $X={1,2,3,4}$ We can still add element in it and it will still be a filter for instance by adding the element ${1}$ from $X={1,2,3,4}$ we can have the filter $F={{1},{1,4},{1,2,4},{1,3,4},{1,2,3,4}}$ This is an Ultrafilter on $X={1,2,3,4}$ as we cannot add any further element from $X={1,2,3,4}$ that satisfies closures on intersection.

There is another collection of sets taken from $X={1,2,3,4}$ Which is the powerset $P={{},{1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}}$ This contain the null element $∅= {}$ so we cannot call this as Ultrafilter. This is not a proper filter according to the article in Wikipedia. In the powerset every intersection of element is again in the powerset again but it contains the null element $∅= {}$ and isn't classified as proper filter.

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    Dec 11, 2022 at 23:21

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