Introduction to filter 1. Basic definition and examples

I would like introducd abstract filters first, explain where it comes from (mathematically instead of historically).

The filters is the dual concept of ideal in lattice. So, where are the ideals in lattice comes from?

It is comes from the equivalence between Boolean Algebra(which is a kind of lattice) and Boolean Ring.

You can view the ideals in lattice is generalized from the ideals in Boolean Ring.

Definition. Let $L$$L$ be a lattice. A non-empty subset $J$$J$ of $L$$L$ is called an ideal if

$$a,b\in J⟹a\vee b\in J,(a\in L,b\in J)\wedge a\le b⟹a\in J$$

In other word, an Ideal $J\subseteq L$$J\subseteq L$ is a non empty down set closed under join.

Categorically speaking, $J$$J$ is such a subcategory of $L$$L$, it is closed under coproduct, and if $\mathrm{\forall }b\in J,{\mathrm{Hom}}_L(a,b)\ne \mathrm{\varnothing }⟹a\in J$$\forall b\in J,\mathrm{Hom}_L(a,b)\ne\empty\implies a\in J$.

Filter is the dual concept of ideal.

Definition. Let $L$$L$ be a lattice. A non-empty subset $\mathcal{F}$$\cal F$ of $L$$L$ is called an filter if $\mathcal{F}$$\cal F$ is an ideal in $L^{\mathrm{op}}$$L^{\mathrm{op}}$.

That is

$$a,b\in \mathcal{F}⟹\mathcal{a}\wedge \mathcal{b}\in \mathcal{F}\mathcal{,}\mathcal{a}\in \mathcal{L}\mathcal{,}\mathcal{b}\in \mathcal{F}\wedge \mathcal{a}\ge \mathcal{b}⟹\mathcal{a}\in \mathcal{F}$$

The aim of this blogs is to introduce the application of filters in topology. So we will consider $L=P(X)$$L=P(X)$​.

In ring theory, we could consider some elements $a_1,...,a_n\in R$$a_1,...,a_n\in R$ and the ideal $(a_1,...,a_n)$$(a_1,...,a_n)$. Similarly, we could consider the concept filter bases, wchih will generate a filter.

Definition. Let $\mathfrak{B}\ne \mathrm{\varnothing }$$\frak B\ne\empty$ be a family of subset of $X$$X$. If $A,B\in \mathfrak{B}⟹\mathrm{\exists }C\in \mathfrak{B},C\subseteq A\cap B$$A,B\in\mathfrak B\implies\exists C\in\mathfrak B,C\subseteq A\cap B$ and $\mathrm{\varnothing }\notin \mathfrak{B}$$\empty\notin\mathfrak B$. Then we call $\mathfrak{B}$$\mathfrak B$ a filter bases. For a filter bases $\mathfrak{B}$$\mathfrak B$, we call $\mathcal{F}:=\{F\subseteq X:\mathrm{\exists }B\in \mathfrak{B},F\supseteq B\}$$\mathcal F:=\{F\subseteq X:\exists B\in\mathfrak B,F\supseteq B\}$ the filter generated by $\mathfrak{B}$$\mathfrak B$​.

Well, as we see, every $\mathcal{F}$$\mathcal F$ is a subset of $P(X)$$P(X)$, hence element in $PP(X)$$PP(X)$. We could define the order between different filters via the order on $PP(X)$$PP(X)$. i.e. ${\mathcal{F}}_1\le {\mathcal{F}}_2⟺{\mathcal{F}}_1\subseteq {\mathcal{F}}_2$$\mathcal F_1\le\mathcal F_2\iff\mathcal F_1\subseteq\mathcal F_2$. Denote the set of all the filter on $X$$X$ as ${\mathcal{F}}_X$$\mathcal F_X$. We will discuss more on ${\mathcal{F}}_X$$\mathcal F_X$ in the future.

Definition. Let $X$$X$ be a topological space. A sequence in $X$$X$ is a function $x:\mathbb{N}\to X$$x:\mathbb N\to X$. A sequence $\{x_n\}$$\{x_n\}$ convergence to $z\in X$$z\in X$ if and only if for every open set $U$$U$ containing $z$$z$, there exisits an $N\in \mathbb{N}$$N\in\mathbb N$ so that if $n\ge N$$n\ge N$ then $x_n\in U$$x_n\in U$. We will write $\{x_n\}\to z$$\{x_n\}\to z$ when $x_n$$x_n$ convergence to $z$$z$.

Remark. $\{x_n\}$$\{x_n\}$ need not to convergence to a unique point. For example, consider $(X,\tau ),|X|\ge 2$$(X,\tau),|X|\ge 2$ with trivial topology.

Consider the category of sequence. Whcih is the coslice category of $\mathbb{N}/\mathrm{Top}$$\mathbb N/\mathrm{Top}$. Here we consider $\mathbb{N}$$\mathbb N$ with discrete topology.

The objects are $x:\mathbb{N}\to X$$x:\N\to X$. The morphisms are $f:X\to Y,x_n⟼f(x_n)$$f:X\to Y,x_n\longmapsto f(x_n)$​.

Now we can consider an example of filter induced by $\{x_n\}$$\{x_n\}$.

Example. Let $x_n$$x_n$ be a sequence in $X$$X$. Let ${\mathcal{F}}_{x_n}$$\mathcal F_{x_{n}}$ be all the $E\subseteq X$$E\subseteq X$ satisfies:

$$\mathrm{\exists }N\ge 1,k\ge N⟹x_k\in E$$

It forms a filter. Let $A,B\subseteq {\mathcal{F}}_{x_n}$$A,B\subseteq \mathcal F_{x_{n}}$. $\mathrm{\exists }N\ge 1,k\ge N⟹x_k\in A$$\exists N\ge 1,k\ge N\implies x_k\in A$ and $\mathrm{\exists }M\ge 1,k\ge M⟹x_k\in B$$\exists M\ge 1,k\ge M\implies x_k\in B$.

Let $\mu =max\{N,M\}$$\mu=\max\{N,M\}$,

$$\mathrm{\exists }\mu \ge 1,k\ge \mu ⟹x_k\in A\cap B$$

Easy to see it is a up set.

The filter ${\mathcal{F}}_{x_n}$$\mathcal F_{x_{n}}$ is generated by $E_k=\{x_k,x_{k+1}...\}$$E_k=\{x_k,x_{k+1}...\}$.

Example. Let $x\in (X,\tau )$$x\in(X,\tau)$. Consider the open neighbourhood of $x$$x$. i.e. $U\in \tau ,x\in U$$U\in\tau,x\in U$. Denote it as ${\tau }_x$$\large\tau_x$.

It is not hard to check that ${\tau }_x$$\large\tau_x$ form a filter bases. Since $\mathrm{\varnothing }\notin {\tau }_x$$\empty\notin\large\tau_x$ and for $U,V\in {\tau }_x$$U,V\in\large\tau_x$, $U\cap V\in {\tau }_x$$U\cap V\in\large\tau_x$ .

The filter generated by ${\tau }_x$$\large\tau_x$ is ${\mathfrak{N}}_x$$\mathfrak N_x$​. We call it neighborhood filter.

Well, what is the connection between this two examples?

Theorem. $x_n$$x_n$ convergence to $x$$x$ if and only if ${\tau }_x\le {\mathcal{F}}_{x_n}$${\large{\tau}_x}\le\mathcal F_{x_{n}}$ .

Proof. Recall the definition.

A sequence $\{x_n\}$$\{x_n\}$ convergence to $z\in X$$z\in X$ if and only if for every open set $U$$U$ containing $z$$z$, there exisits an $N\in \mathbb{N}$$N\in\mathbb N$ so that if $n\ge N$$n\ge N$ then $x_n\in U$$x_n\in U$.

That is, $\mathrm{\forall }U\in {\tau }_x$$\forall U\in\large\tau_x$, there exist $E_N\subseteq U$$E_N\subseteq U$. Hence $U$$U$ in the upper set of $E_N$$E_N$. By definition of filter bases, $U\in {\mathcal{F}}_{x_n}$$U\in\mathcal F_{x_{n}}$.

Hence we proved ${\tau }_x\le {\mathcal{F}}_{x_n}$${\large{\tau}_x}\le\mathcal F_{x_{n}}$. $◻$$\square$​

Remark. ${\tau }_x\le {\mathcal{F}}_{x_n}$${\large{\tau}_x}\le\mathcal F_{x_{n}}$ means that ${\mathcal{F}}_{x_n}$$\mathcal F_{x_n}$ is in the upper set of ${\tau }_x$$\tau_x$. And we know that $\uparrow {\tau }_x$$\uparrow\tau_x$ is a representation of ${\mathrm{Hom}}_P({\tau }_x,-)$$\mathrm{Hom}_P(\tau_x,-)$.

In general, we can define the convergence of filter bases as follows.

Definition. Let $\mathfrak{B}$$\mathfrak B$ be a filter bases over $X$$X$. We say $\mathfrak{B}$$\mathfrak B$ convergence to $x\in X$$x\in X$, if every $U\in {\tau }_x$$U\in\large\tau_x$ contain a $F\in \mathfrak{B}$$F\in\mathfrak B$.

Let $\mathcal{F}$$\mathcal F$ be the filter generated by $\mathfrak{B}$$\mathfrak B$. The filter bases $\mathfrak{B}$$\mathfrak B$ convergence to $x\in X$$x\in X$ implies ${\tau }_x\le \mathcal{F}$${\large{\tau}_x}\le\mathcal F$​.

By definition, ${\tau }_x$$\large\tau_x$ convergence to $x$$x$.

Proposition. Let $\mathfrak{B}$$\mathfrak B$ be a filter bases, then $f\mathfrak{B}$$f\mathfrak B$ is filter bases as well.

Proof Recall the definition.

Let $\mathfrak{B}\ne \mathrm{\varnothing }$$\frak B\ne\empty$ be a family of subset of $X$$X$. If $A,B\in \mathfrak{B}⟹\mathrm{\exists }C\in \mathfrak{B},C\subseteq A\cap B$$A,B\in\mathfrak B\implies\exists C\in\mathfrak B,C\subseteq A\cap B$ and $\mathrm{\varnothing }\notin \mathfrak{B}$$\empty\notin\mathfrak B$. Then we call $\mathfrak{B}$$\mathfrak B$ a filter bases. For a filter bases $\mathfrak{B}$$\mathfrak B$, we call $\mathcal{F}:=\{F\subseteq X:\mathrm{\exists }B\in \mathfrak{B},F\supseteq B\}$$\mathcal F:=\{F\subseteq X:\exists B\in\mathfrak B,F\supseteq B\}$ the filter generated by $\mathfrak{B}$$\mathfrak B$​.

and $f(C)\subseteq f(A\cap B)\subseteq f(A)\cap f(B)$$f(C)\subseteq f(A\cap B)\subseteq f(A)\cap f(B)$. $◻$$\square$

Let me introduce a way to deduce filters.

Let $f:P(X)\to 2$$f:P(X)\to 2$ be a semilattice homomorphism. That is, $f(A\cap B)=f(A)\wedge f(B)$$f(A\cap B)=f(A)\wedge f(B)$. Then $f^{-1}(1)$$f^{-1}(1)$ is a filter.

Proof. Let $U,V\in f^{-1}(1)$$U,V\in f^{-1}(1)$. Then $U\cap V\in f^{-1}(1)$$U\cap V\in f^{-1}(1)$. Since $f(U\cap V)=f(U)\wedge f(V)=1\wedge 1=1.$$f(U\cap V)=f(U)\wedge f(V)=1\wedge 1=1.$

It is upper set follows from $f$$f$ preserve order.

$$U\subseteq V⟹U\cup V=V⟹f(U)\cup f(V)=f(V)⟹f(U)\le f(V)◻$$

Conversely, given a filter $\mathcal{F}$$\mathcal F$ on $X$$X$, we can define a function as follows:

$$\begin{array}{c}f(U)=\{\begin{array}{l}1\text{ if }U\in \mathcal{F}\\ 0\text{ otherwise}\end{array}\end{array}$$

It is a semilattice homomorphism.

Proof.

Since if $U,V\in \mathcal{F},f(U\cap V)=1=f(U)\cap f(V)$$U,V\in\mathcal F,f(U\cap V)=1=f(U)\cap f(V)$.

If $U\in \mathcal{F},V\notin \mathcal{F}⟹U\cap V\notin \mathcal{F}$$U\in\mathcal F,V\notin\mathcal F\implies U\cap V\notin\mathcal F$. Otherwise $U\cap V\subseteq V⟹V\in \mathcal{F}$$U\cap V\subseteq V\implies V\in\mathcal F$​.

If $U,V\notin \mathcal{F}$$U,V\notin\mathcal F$, then $f(U\cap V)=0$$f(U\cap V)=0$ since $U\cap V\notin \mathcal{F}$$U\cap V\notin\mathcal F$ for same reason. Then $f(U\cap V)=0=f(U)\cap f(V).◻$$f(U\cap V)=0=f(U)\cap f(V).\quad\square$