Intuition behind filter on a set

Solution 1:

Generally each filter encodes a possible meaning of what it means for a subset of $X$ to be "large enough" or "contain enough points" for something.

In addition to Asaf's examples, other good filters to keep in mind are

1. The principal filter of $x\in X$, which is the set of all subsets of $X$ that contain $x$. Here "large enough" is almost trivial, being large enough just means to contain $x$.

2. The neighborhood filter of $x$, when $X$ is a topological space. This is the set of all subsets of $X$ that have $x$ as an interior point.

These example show that being "large" or (in Asaf's words) including "most" elements can be very context dependent and doesn't need to correspond to "most" in an objective sense.

The neighborhood filter is particularly important because it is the prototype of the use of filters to generalize the notion of limits. Filters are one possible answer to what kind of thing does "$x\to 5$" in $\lim\limits_{x\to 5} f(x)$ stand for.

For example, when $\mathcal F$ is a filter in $X$ and $f$ is some function defined on $X$, we can define $$\lim_{x\to\mathcal F} f(x)=y \text{ means that } \forall \varepsilon>0\, \exists F\in\mathcal F\, \forall x\in F : |f(x)-y|<\varepsilon$$ If we choose $\mathcal F$ to be the punctured-neighborhood filter at $5$ gives us the usual limit for $x\to 5$. But there are other filters that give one-sided limits ($x\to 5^+$ corresponds to the filter of all sets that contain $(5,5+\delta]$ for some $\delta>0$) or limits involving infinity ($x\to-\infty$ corresponds to the filter of all sets that contain $(-\infty,y]$ for some $y\in\mathbb R$), and so forth, which can now be considered instances of a single uniform concept.

Solution 2:

Filters come to model the notion of "large sets", or subsets which include "most" of the elements of $X$.

1. Most elements of $X$ are in $X$; and most elements are not in $\varnothing$.
2. If most elements are in $A$ and most elements are in $B$, then most elements are in $A\cap B$.
3. If most elements are in $A$, and $A\subseteq B$, then of course that most elements are in $B$.

Examples for "most" can be seen in the canonical examples, the cofinite filter, namely $\mathcal F=\{A\subseteq X\mid X\setminus A\text{ is finite}\}$. Then every $A\in\mathcal F$ includes most of the elements of $X$ in this aspect.

Similarly, $\{A\subseteq[0,1]\mid m(A)=1\}$ where $m$ is the Lebesgue measure, also gives us a filter on $[0,1]$ which give us subsets which are almost everything in terms of measure.