Convergence: Nets and Filters

Suppose $x$ is eventually in $S\subseteq X$, and choose $d_0$ such that $d\ge d_0$ implies $x_d\in S$.

Choose $e_0$ from [A1] for this $d_0$; then $e\ge e_0$ gives $y_e=x_{\varphi (e)}\in S$. Thus $y$ is eventually in $S$.

If $x$ converges to $p$, apply step 2.1 to each neighbourhood of $p$ using [A2]; then $y$ is eventually in every such neighbourhood and converges to $p$.

Suppose $p$ is a cluster point. Let $E=\{(d,N):N\in \mathcal{N}(p),d\in D,x_d\in N\}$, ordered by $(d,N)⪯(e,M)$ when $d\le e$ and $M\subseteq N$.

Conversely, suppose a subnet $y_e=x_{\varphi (e)}$ converges to $p$. Given a neighbourhood $N$ and $d\in D$, choose $e_0$ after which $y$ lies in $N$ and choose $e_1$ after which $\varphi (e)\ge d$; a common upper bound $e$ of $e_0,e_1$ gives $\varphi (e)\ge d$ and $x_{\varphi (e)}=y_e\in N$. Hence $x$ is frequently in $N$.

The set $E$ is directed: for $(d,N),(e,M)\in E$, take $h\ge d,e$ in $D$; frequent membership in $N\cap M$ gives $k\ge h$ with $x_k\in N\cap M$, and $(k,N\cap M)$ is above both pairs.

Put $y_{(d,N)}=x_d$ and $\varphi (d,N)=d$. For every $d_0\in D$, the pair $(d_0,X)$ lies in $E$, and every later pair has first coordinate at least $d_0$. Thus $\varphi$ is eventually cofinal and $y$ is a subnet of $x$.

For a neighbourhood $N$ of $p$, choose $(d,N)\in E$ using frequent membership in $N$. Every pair later than it has second coordinate contained in $N$, hence its $y$-value lies in $N$. Thus $y\to p$.

Steps 1.1 and 2.1--2.3 construct a convergent subnet from a cluster point, and step 1.2 gives the converse.

Suppose $p\in \bar{A}$. Let $E=\{(N,a):N\in \mathcal{N}(p),a\in N\cap A\}$, ordered by $(N,a)⪯(M,b)$ when $M\subseteq N$, and put $x_{(N,a)}=a$.

Conversely, if a net $x$ in $A$ converges to $p$, every neighbourhood $N$ of $p$ contains some eventual value $x_d\in A$, so $N\cap A\mathrm{\ne }\mathrm{\varnothing }$ and $p\in \bar{A}$.

The index set is directed: for $(N,a),(M,b)$, the set $N\cap M$ is a neighbourhood and meets $A$; for $c\in (N\cap M)\cap A$, the pair $(N\cap M,c)$ is above both.

Given a neighbourhood $N$ of $p$, choose $(N,a)\in E$. Every later pair has its second coordinate in a subset of $N$, so $x$ is eventually in $N$ and therefore converges to $p$.

Steps 1.1--2.1 construct the required net and step 1.2 proves the converse.

If $f$ is continuous at $p$ and $x_d\to p$, then for every neighbourhood $V$ of $f(p)$ the net is eventually in $f^{-1}[V]$ by [A1], hence $f(x_d)$ is eventually in $V$ and converges to $f(p)$.

Conversely, assume every net converging to $p$ has image converging to $f(p)$, and assume for a contradiction that $f$ is not continuous at $p$. Then some neighbourhood $V$ of $f(p)$ has $f^{-1}[V]$ not a neighbourhood of $p$.

Put $A=X\setminus f^{-1}[V]$. Every neighbourhood of $p$ meets $A$, for otherwise it would be contained in $f^{-1}[V]$; hence $p\in \bar{A}$ and [A2] gives a net $x_d$ in $A$ converging to $p$.

Every $f(x_d)$ lies outside $V$, so its image net is not eventually in the neighbourhood $V$ of $f(p)$ and cannot converge to $f(p)$, contradicting the assumption of step 1.2.

Therefore $f$ is continuous at $p$; together with step 1.1 this proves the equivalence.

Suppose $X$ is Hausdorff and a net converges to both $p$ and $q$. If $p\mathrm{\ne }q$, take disjoint neighbourhoods $U$ of $p$ and $V$ of $q$; the net is eventually in both, and directedness supplies an index after both thresholds, whose value would lie in $U\cap V$.

Conversely, suppose $X$ is not Hausdorff. Choose distinct $p,q$ for which every neighbourhood of $p$ meets every neighbourhood of $q$, and let $E=\{(U,V,z):U\in \mathcal{N}(p),V\in \mathcal{N}(q),z\in U\cap V\}$, ordered by reverse inclusion in the first two coordinates.

Thus $p=q$, so every net has at most one limit.

The set $E$ is directed: intersect the first two neighbourhood coordinates of two triples and choose a point in their intersection; the resulting triple is above both. The net sending $(U,V,z)$ to $z$ is eventually in every neighbourhood of $p$ and every neighbourhood of $q$, hence converges to both distinct points.

Therefore uniqueness of all net limits forces $X$ to be Hausdorff, and the two implications prove the result.

For every neighbourhood $N$ of $p$, $x$ is eventually in $N$ exactly when $N\in {\mathcal{F}}_x$ by [A1]. Thus the two convergence conditions in [A2] are equivalent.

For every neighbourhood $N$ of $p$, $x$ is frequently in $N$ exactly when $N$ meets every tail $T_d$: a point in $N\cap T_d$ is a value $x_e\in N$ with $e\ge d$.

If $N$ meets every tail, it meets every member of ${\mathcal{F}}_x$, since each such member contains a tail; conversely every tail belongs to ${\mathcal{F}}_x$. Hence the two cluster-point conditions in [A2] are equivalent.

If a neighbourhood $N$ of $p$ belongs to $\mathcal{F}$, choose $x\in N$; then $(N,x)$ is an index, and every later $(B,y)$ has $B\subseteq N$, hence $y\in N$. Thus filter convergence implies convergence of the derived net.

If the derived net is eventually in $N$, take a threshold $(A,x)$. Applying eventuality to indices $(A,y)$ with $y\in A$ gives $A\subseteq N$; upward closure of the filter gives $N\in \mathcal{F}$. Thus convergence is equivalent.

The derived net is frequently in $N$ exactly when every $A\in \mathcal{F}$ meets $N$: after $(A,x)$ a point of $A\cap N$ supplies a later index, and conversely frequent membership after $(A,x)$ supplies such a point. Therefore its cluster points are exactly those of $\mathcal{F}$.

Apply [L1] to the given net.

Apply [L2] to the given filter.

These are precisely the two asserted correspondences.

By [A1], universality of $x$ says exactly that its tail filter contains $S$ or $X\setminus S$ for every $S\subseteq X$. By [A2], this is exactly ultrafilterhood.

If $\mathcal{U}$ is an ultrafilter and $S\subseteq X$, [A2] gives $S\in \mathcal{U}$ or $X\setminus S\in \mathcal{U}$. In the first case an index $(S,a)$ exists and every later value lies in $S$ by [A3]; the second case is identical.

Thus the derived net of an ultrafilter is universal, completing both assertions.

Assume for a contradiction that $\mathcal{U}$ does not converge to $p$. Then some neighbourhood $N$ of $p$ is not in $\mathcal{U}$.

By [A2], $X\setminus N\in \mathcal{U}$. But $N$ must meet every member of $\mathcal{U}$ by clusterhood, whereas $N\cap (X\setminus N)=\mathrm{\varnothing }$.

This contradiction proves $\mathcal{U}\to p$.

Suppose $X$ is compact and $\mathcal{F}$ is a filter. The closed family $\{\bar{A}:A\in \mathcal{F}\}$ has the finite-intersection property, because a finite intersection of members of $\mathcal{F}$ is nonempty and is contained in the corresponding intersection of closures. By [L1], choose $p\in {\bigcap }_{A\in \mathcal{F}}\bar{A}$.

If every filter has a cluster point, apply this to a net's tail filter and use [L3]; hence 4 implies 2. By [L2], conditions 2 and 3 are equivalent.

Conversely, if every net has a cluster point and $\mathcal{F}$ is a filter, its derived net has a cluster point, which is also a cluster point of $\mathcal{F}$ by [L3]. Hence 2 implies 4.

Condition 4 implies 5 because an ultrafilter is a filter and [L4] turns its cluster point into a limit.

Suppose every ultrafilter converges and let $\mathcal{C}$ be a family of closed subsets of $X$ with the finite-intersection property. Clause 2 of [L1] gives a filter containing $\mathcal{C}$, and [L4] extends it to an ultrafilter $\mathcal{U}$.

Every neighbourhood of $p$ meets every $A\in \mathcal{F}$, since $p\in \bar{A}$; thus $p$ is a cluster point of $\mathcal{F}$. Hence 1 implies 4.

Let $p$ be a limit of $\mathcal{U}$. For $C\in \mathcal{C}$, every neighbourhood of $p$ belongs to $\mathcal{U}$ and meets $C\in \mathcal{U}$; therefore $p\in \bar{C}=C$. Thus $\bigcap \mathcal{C}\mathrm{\ne }\mathrm{\varnothing }$, and [L1] gives compactness.

The implications in steps 2.1, 1.2, 1.3, 1.4 and 2.2 establish all five equivalences.

Assume for a contradiction that $x$ does not converge to $p$. Then some neighbourhood $N$ of $p$ is not an eventual set for $x$.

By universality, $x$ is eventually in $X\setminus N$. This contradicts frequent membership in $N$, since an index after both thresholds would lie in $N\cap (X\setminus N)$.

Therefore $x$ converges to $p$.

Let $S\subseteq Y$. By [A1], $x$ is eventually in $f^{-1}[S]$ or in its complement $f^{-1}[Y\setminus S]$; respectively, $f(x)$ is eventually in $S$ or in $Y\setminus S$.

Thus $f(x)$ is universal.

If $f$ is continuous and $x\to p$, the second assertion is [L1].

Choose an ultrafilter $\mathcal{U}\supseteq {\mathcal{F}}_x$ by [L1]. Let $E=\{(d,A):d\in D,A\in \mathcal{U},x_d\in A\}$, ordered by $(d,A)⪯(e,B)$ when $d\le e$ and $B\subseteq A$, and put $y_{(d,A)}=x_d$.

The set $E$ is directed. Given $(d,A),(e,B)$, choose $h\ge d,e$. Since $A\cap B$ and the tail $T_h$ belong to $\mathcal{U}$, their intersection is nonempty; choose an index $k\ge h$ with $x_k\in A\cap B$. Then $(k,A\cap B)$ is above both pairs.

The map $\varphi (d,A)=d$ is eventually cofinal: $(d_0,X)$ is an index for every $d_0$, and every later index has first coordinate at least $d_0$. Thus $y$ is a subnet of $x$.

For $S\subseteq X$, [L2] gives $S\in \mathcal{U}$ or $X\setminus S\in \mathcal{U}$. In the first case choose any $d_0\in D$. Since $S\cap T_{d_0}\in \mathcal{U}$, choose $j\ge d_0$ with $x_j\in S$. Then $(j,S)\in E$, and every later value lies in $S$. The complementary case is identical. Thus $y$ is universal.

The constructed $y$ is a universal subnet of $x$.

If $X$ is compact, a universal net has a cluster point by [L1], hence converges by [L2].

Conversely, suppose every universal net converges. Every net has a universal subnet by [L2], which then converges; its limit is a cluster point of the original net by [L3]. Thus every net has a cluster point.

By [L1], this makes $X$ compact.

For every $i\in I$, the projection ${\pi }_i$ is continuous, so ${\pi }_i(x_d)$ is universal by [L1] and converges in compact $X_i$ by [L2]. Its limit $p_i$ is unique by [L3].

The uniqueness in step 1.1 defines a point $p\in {\prod }_{i\in I}{X}_i$, namely the function $i\mapsto p_i$, rather than choosing a family of limits.

Let $N$ be a neighbourhood of $p$ in $P$. By [L4], it contains a basic product neighbourhood restricting a finite set $J\subseteq I$; for each $i\in J$, the coordinate net is eventually in its prescribed neighbourhood of $p_i$. Directedness supplies one index after the finitely many thresholds, and after it $x_d\in N$. Thus $x_d\to p$.

Every universal net in $P$ converges by step 2.2. The converse direction of [L2] therefore makes $P$ compact.

Under countable choice, [L1] is exactly the defining equality for a first countable space to be Fréchet–Urysohn.

Now suppose $X$ is Fréchet–Urysohn and $C$ is sequentially closed. Then $\mathrm{seqcl}(C)=C$, because the constant sequence gives $C\subseteq \mathrm{seqcl}(C)$ and sequential closedness gives the reverse inclusion.

Fréchet–Urysohnness gives $\bar{C}=\mathrm{seqcl}(C)=C$, so $C$ is closed. Therefore $X$ is sequential.

Put $\varphi (0)=0$ and $\varphi (k)=k-1$ for $k\ge 1$, and let $y_k=x_{\varphi (k)}$. For every $n$, all $k\ge n+1$ satisfy $\varphi (k)=k-1\ge n$, so $\varphi$ is eventually cofinal and $y$ is a subnet of $x$.

The subnet has $y_0=y_1=0$. Every subsequence of the injective identity sequence is injective by [A2], so $y$ cannot be a subsequence of $x$.

Thus the stated universal claim is false.