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 $(N,a)\preceq(M,b)$ 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 $A$.

$\operatorname{Fin}(I)$ is nonempty because it contains $\varnothing$, 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 $|\sum_{i\in F}a_i|\le C$ 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 $\sum_{i\in P}a_i =\sum_{i\in F_0\cup P}a_i-\sum_{i\in F_0}a_i \le C+\left|\sum_{i\in F_0}a_i\right|.$ 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 $\varepsilon>0$, choose a finite initial segment $F_0$ whose remaining absolute series sum is below $\varepsilon$. For every finite $F\supseteq F_0$, $\left|\sum_{i\in F}a_i-s\right| \le \sum_{i\in S\setminus F}|a_i| <\varepsilon.$ Indices outside $S$ contribute zero, so the finite-subset net converges to $s$.

For each $n\ge1$, the sets $\{i\notin F_0:a_i^+\ge1/n\}$ and $\{i\notin F_0:a_i^-\ge1/n\}$ are finite, since a finite subset with more than $nC'$ members would have sum exceeding the bound $C'$. 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 $\{k\}\cap T_{k+1}=\varnothing$; 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 $\infty$ contains some $U_N$, so upward closure gives $\mathcal U^X\to\infty$.

It is free: if $\{x\}\in\mathcal U^X$, then either $x=\infty$ 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 $2\in\overline A$, 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 $\infty\in\overline A$. If every spoke met $A$ only finitely, define $f(n)= \begin{cases} 0,&\{m:(n,m)\in A\}=\varnothing,\\ 1+\max\{m:(n,m)\in A\},&\text{otherwise}. \end{cases}$ 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 $\infty$. 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 $\infty$. 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 $\infty$ meets $A$, so $\infty\in\overline A$ by [L1]. No sequence in $A$ converges to $\infty$: 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 $x_n\in\overline C$, 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 $\infty\in\overline C$, 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 $\infty$ disjoint from $C$, contradicting $\infty\in\overline C$.

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 $\infty$, so sequential closedness puts $\infty$ 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.