Convergence: Nets and Filters: Examples and Counterexamples

Let $E=\{(N,a):N\in \mathcal{N}(p),a\in N\cap A\}$ and order it by $(N,a)⪯(M,b)$ when $M\subseteq N$.

For two indices, [L2] and [L1] give $c\in (N\cap M)\cap A$; $(N\cap M,c)$ is above both. Thus $E$ is directed.

The net $x_{(N,a)}=a$ is eventually in every neighbourhood $N$ of $p$, since any pair with first coordinate $N$ is a threshold. Hence $x\to p$.

This is the asserted net in $A$.

$\mathrm{Fin}(I)$ is nonempty because it contains $\mathrm{\varnothing }$, and it is directed because $F\cup G$ is a finite upper bound of $F$ and $G$.

Therefore $F\mapsto s_F$ is a net. If $F\subseteq G$, then $s_G=s_F+{\sum }_{i\in G\setminus F}{a}_i$, so later values add only terms not already counted.

Thus the displayed finite partial sums form the announced net, and its convergence is a definition of unordered summability.

Suppose the finite-subset net converges to $L$. There are a finite $F_0\subseteq I$ and $C>0$ such that $\mathrm{\mid }{\sum }_{i\in F}{a}_i\mathrm{\mid }\le C$ for every finite $F\supseteq F_0$. If $P\subseteq I\setminus F_0$ is finite and all $a_i$ for $i\in P$ are positive, then ${\sum }_{i\in P}{a}_i={\sum }_{i\in F_0\cup P}{a}_i-{\sum }_{i\in F_0}{a}_i\le C+\mid {\sum }_{i\in F_0}{a}_i\mid \mathrm{.}$ The same argument applied to finite sets of negative terms bounds their absolute-value sums.

Conversely, let an enumeration of $S$ have absolutely convergent series sum $s$. Given $\epsilon >0$, choose a finite initial segment $F_0$ whose remaining absolute series sum is below $\epsilon$. For every finite $F\supseteq F_0$, $\mid {\sum }_{i\in F}{a}_i-s\mid \le {\sum }_{i\in S\setminus F}\mathrm{\mid }{a}_i\mathrm{\mid }<\epsilon \mathrm{.}$ Indices outside $S$ contribute zero, so the finite-subset net converges to $s$.

For each $n\ge 1$, the sets $\{i\mathrm{\notin }{F}_0:a_i^{+}\ge 1\mathrm{/}n\}$ and $\{i\mathrm{\notin }{F}_0:a_i^{-}\ge 1\mathrm{/}n\}$ are finite, since a finite subset with more than $n{C}^{\mathrm{\prime }}$ members would have sum exceeding the bound $C^{\mathrm{\prime }}$. Every nonzero real lies in one of these level sets for some $n$ by [L4], so [L1] makes $S$ at most countable.

With any enumeration of $S$, the positive and negative partial sums are bounded by step 1.1, hence converge by [L2]. Thus the series of absolute values converges by [L3], so the enumerated nonzero terms form an absolutely convergent series.

Any two infinite enumerations differ by a bijective rearrangement, so [L2] gives the same sum; finite enumerations give the same finite-set sum by [L5]. This proves both directions and enumeration independence.

Choose an ultrafilter $\mathcal{U}$ extending the tail filter. It contains every $T_N$ and contains no singleton, since $\{k\}\cap T_{k+1}=\mathrm{\varnothing }$; thus it is free.

Put ${\mathcal{U}}^X=\{B\subseteq X:B\cap \mathbb{N}\in \mathcal{U}\}$. The filter axioms transfer through intersection with $\mathbb{N}$, so this is a filter on $X$. For every $B\subseteq X$, [L5] applied to $B\cap \mathbb{N}$ shows that ${\mathcal{U}}^X$ contains $B$ or $X\setminus B$; hence ${\mathcal{U}}^X$ is an ultrafilter.

Every basic neighbourhood $U_N$ has $U_N\cap \mathbb{N}=T_N\in \mathcal{U}$, hence $U_N\in {\mathcal{U}}^X$. Every neighbourhood of $\mathrm{\infty }$ contains some $U_N$, so upward closure gives ${\mathcal{U}}^X\to \mathrm{\infty }$.

It is free: if $\{x\}\in {\mathcal{U}}^X$, then either $x=\mathrm{\infty }$ and its intersection with $\mathbb{N}$ is empty, or $x\in \mathbb{N}$ and $\{x\}\in \mathcal{U}$, both impossible. Thus this supplies the claimed free ultrafilter and its convergence.

Each two-point discrete factor is compact and Hausdorff, so the cube is compact under the ultrafilter lemma by [L3]; then [L2] gives a convergent subnet of $(F_n)$.

Assume for a contradiction that $F_{n_j}$ is a convergent subsequence. Define $r\in D$ by $r_{n_j}=0$ for even $j$ and $r_{n_j}=1$ for odd $j$, assigning $0$ elsewhere.

The $r$-coordinate of $F_{n_j}$ alternates $0,1$, so it does not converge in the discrete two-point factor. By [L4], a convergent product net has convergent coordinate nets, contradiction.

Hence no convergent subsequence exists, while step 1.1 supplies a convergent subnet.

Every neighbourhood $N$ of $2$ has at most countable complement, so it meets the uncountable set $A$. Hence $2\in \bar{A}$, and [L4] supplies a net in $A$ converging to $2$.

Let $(a_n)$ be a sequence in $A$. Its range is at most countable and omits $2$, so $\mathbb{R}\setminus \{a_n:n\in \mathbb{N}\}$ is a neighbourhood of $2$ containing none of its terms. Thus $(a_n)$ does not converge to $2$.

The net from step 1.1 detects the closure point, whereas no sequence in $A$ does.

Suppose $\mathrm{\infty }\in \bar{A}$. If every spoke met $A$ only finitely, define $f(n)=\{\begin{array}{ll}0,& \{m:(n,m)\in A\}=\mathrm{\varnothing },\\ 1+\max \{m:(n,m)\in A\},& \text{otherwise}\mathrm{.}\end{array}$ This is a canonically defined function by [L1], and the neighbourhood containing on spoke $n$ exactly the points $(n,m)$ with $m\ge f(n)$ misses $A$, a contradiction. Hence one spoke meets $A$ infinitely.

Suppose $(B_k)$ were a countable neighbourhood base at $\mathrm{\infty }$. For each $k,n$, let $f_k(n)$ be the least threshold such that $(n,m)\in B_k$ for every $m\ge f_k(n)$; it exists and is unique by [L1]. Form the neighbourhood whose threshold on spoke $k$ is $g(k)=f_k(k)+1$.

On the infinite spoke supplied by step 1.1, recursion and least elements from [L1] list the second coordinates increasingly. The resulting sequence in $A$ is eventually beyond every threshold on that spoke, hence converges to $\mathrm{\infty }$. Isolated closure points already lie in $A$, so $S_{\omega }$ is Fréchet–Urysohn.

The point $(k,f_k(k))$ lies in $B_k$ but not in this neighbourhood, so no $B_k$ is contained in it. This contradicts the base property.

Therefore the sequential fan is Fréchet–Urysohn and not first countable.

Every neighbourhood of $\mathrm{\infty }$ meets $A$, so $\mathrm{\infty }\in \bar{A}$ by [L1]. No sequence in $A$ converges to $\mathrm{\infty }$: if it visits a row infinitely often, a neighbourhood omitting that row defeats convergence. If it visits every row finitely, use [L2] to put the threshold on each visited row one above the maximum selected second coordinate, and threshold $0$ on every unvisited row. The resulting neighbourhood omits the whole sequence.

Let $C$ be sequentially closed. An isolated closure point lies in $C$. If $x_n\in \bar{C}$, then $C$ meets its $n$-th row arbitrarily far out; recursion and least elements from [L2] give a sequence of row points in $C$ converging to $x_n$, so $x_n\in C$. If $\mathrm{\infty }\in \bar{C}$, then infinitely many $x_n$ lie in $C$: otherwise omit the finitely many rows whose centres lie in $C$. In every remaining row, the preceding conclusion shows that $C$ has only finitely many points; using their maximum as in [L2] gives a canonical tail disjoint from $C$. These tails form a neighbourhood of $\mathrm{\infty }$ disjoint from $C$, contradicting $\mathrm{\infty }\in \bar{C}$.

Hence $S_2$ is not Fréchet–Urysohn.

The indices $n$ with $x_n\in C$ form an infinite subset of $\mathbb{N}$; list them increasingly using [L2]. The resulting sequence of row centres converges to $\mathrm{\infty }$, so sequential closedness puts $\mathrm{\infty }$ in $C$. Thus every sequentially closed $C$ contains all its closure points and is closed. Therefore $S_2$ is sequential.

The two conclusions prove the example.