This theory continues Topology_ZF_4 with results relating the net and filter convergence notions introduced there.
In this section we show that filters do not generalize nets, but still nets and filters are in a way equivalent as far as convergence is considered.
Let's build now a net from a filter, such that both converge to the same points.
definition
\( \mathfrak{F} \text{ is a filter on } \bigcup \mathfrak{F} \Longrightarrow \text{Net}(\mathfrak{F} ) \equiv \) \( \langle \{\langle A,\text{fst}(A)\rangle .\ A\in \{\langle x,F\rangle \in (\bigcup \mathfrak{F} )\times \mathfrak{F} .\ x\in F\}\},\{\langle A,B\rangle \in \{\langle x,F\rangle \in (\bigcup \mathfrak{F} )\times \mathfrak{F} .\ x\in F\}\times \{\langle x,F\rangle \in (\bigcup \mathfrak{F} )\times \mathfrak{F} .\ x\in F\}.\ \text{snd}(B)\subseteq \text{snd}(A)\}\rangle \)
Net of a filter is indeed a net.
theorem net_of_filter_is_net:
assumes \( \mathfrak{F} \text{ is a filter on } X \)
shows \( (\text{Net}(\mathfrak{F} )) \text{ is a net on } X \)proofIf a filter converges to some point then its net converges to the same point.
theorem (in topology0) filter_conver_net_of_filter_conver:
assumes \( \mathfrak{F} \text{ is a filter on } \bigcup T \) and \( \mathfrak{F} \rightarrow _F x \)
shows \( (\text{Net}(\mathfrak{F} )) \rightarrow _N x \)proofIf a net converges to a point, then a filter also converges to a point.
theorem (in topology0) net_of_filter_conver_filter_conver:
assumes \( \mathfrak{F} \text{ is a filter on }\bigcup T \) and \( (\text{Net}(\mathfrak{F} )) \rightarrow _N x \)
shows \( \mathfrak{F} \rightarrow _F x \)proofA filter converges to a point if and only if its net converges to the point.
theorem (in topology0) filter_conver_iff_net_of_filter_conver:
assumes \( \mathfrak{F} \text{ is a filter on }\bigcup T \)
shows \( (\mathfrak{F} \rightarrow _F x) \longleftrightarrow ((\text{Net}(\mathfrak{F} )) \rightarrow _N x) \) using filter_conver_net_of_filter_conver, net_of_filter_conver_filter_conver, assmsThe previous result states that, when considering convergence, the filters do not generalize nets. When considering a filter, there is always a net that converges to the same points of the original filter.
Now we see that with nets, results come naturally applying the axiom of choice; but with filters, the results come, may be less natural, but with no choice. The reason is that \( \text{Net}(\mathfrak{F} ) \) is a net that doesn't come into our attention as a first choice; maybe because we restrict ourselves to the anti-symmetry property of orders without realizing that a directed set is not an order.
The following results will state that filters are not just a subclass of nets, but that nets and filters are equivalent on convergence: for every filter there is a net converging to the same points, and also, for every net there is a filter converging to the same points.
definition
\( (N \text{ is a net on } X) \Longrightarrow Filter N.\ .\ X \equiv \{A\in Pow(X).\ \exists D\in \{\{\text{fst}(N)\text{snd}(s).\ s\in \{s\in domain(\text{fst}(N))\times domain(\text{fst}(N)).\ s\in \text{snd}(N) \wedge \text{fst}(s)=t0\}\}.\ t0\in domain(\text{fst}(N))\}.\ D\subseteq A\} \)
Filter of a net is indeed a filter
theorem filter_of_net_is_filter:
assumes \( N \text{ is a net on } X \)
shows \( (Filter N.\ .\ X) \text{ is a filter on } X \) and \( \{\{\text{fst}(N)\text{snd}(s).\ s\in \{s\in domain(\text{fst}(N))\times domain(\text{fst}(N)).\ s\in \text{snd}(N) \wedge \text{fst}(s)=t0\}\}.\ t0\in domain(\text{fst}(N))\} \text{ is a base filter } (Filter N.\ .\ X) \)proofConvergence of a net implies the convergence of the corresponding filter.
theorem (in topology0) net_conver_filter_of_net_conver:
assumes \( N \text{ is a net on } \bigcup T \) and \( N \rightarrow _N x \)
shows \( (Filter N.\ .\ (\bigcup T)) \rightarrow _F x \)proofConvergence of a filter corresponding to a net implies convergence of the net.
theorem (in topology0) filter_of_net_conver_net_conver:
assumes \( N \text{ is a net on } \bigcup T \) and \( (Filter N.\ .\ (\bigcup T)) \rightarrow _F x \)
shows \( N \rightarrow _N x \)proofFilter of net converges to a point \(x\) if and only the net converges to \(x\).
theorem (in topology0) filter_of_net_conv_iff_net_conv:
assumes \( N \text{ is a net on } \bigcup T \)
shows \( ((Filter N.\ .\ (\bigcup T)) \rightarrow _F x) \longleftrightarrow (N \rightarrow _N x) \) using assms, filter_of_net_conver_net_conver, net_conver_filter_of_net_converWe know now that filters and nets are the same thing, when working convergence of topological spaces. Sometimes, the nature of filters makes it easier to generalize them as follows.
Instead of considering all subsets of some set \(X\), we can consider only open sets (we get an open filter) or closed sets (we get a closed filter). There are many more useful examples that characterize topological properties.
This type of generalization cannot be done with nets.
Also a filter can give us a topology in the following way:
theorem top_of_filter:
assumes \( \mathfrak{F} \text{ is a filter on } \bigcup \mathfrak{F} \)
shows \( (\mathfrak{F} \cup \{0\}) \text{ is a topology } \)proofWe can use topology0 locale with filters.
lemma topology0_filter:
assumes \( \mathfrak{F} \text{ is a filter on } \bigcup \mathfrak{F} \)
shows \( topology0(\mathfrak{F} \cup \{0\}) \) using top_of_filter, topology0_def, assmsThe next abbreviation introduces notation where we want to specify the space where the filter convergence takes place.
abbreviation
\( \mathfrak{F} \rightarrow _F x\text{ in }T \equiv topology0.\ \text{FilterConverges}(T,\mathfrak{F} ,x) \)
The next abbreviation introduces notation where we want to specify the space where the net convergence takes place.
abbreviation
\( N \rightarrow _N x\text{ in }T \equiv topology0.\ \text{NetConverges}(T,N,x) \)
Each point of the union of a filter is a limit of that filter.
lemma lim_filter_top_of_filter:
assumes \( \mathfrak{F} \text{ is a filter on } \bigcup \mathfrak{F} \) and \( x\in \bigcup \mathfrak{F} \)
shows \( \mathfrak{F} \rightarrow _F x\text{ in }(\mathfrak{F} \cup \{0\}) \)proofassumes \( \forall x\in X.\ b(x) \in Y \)
shows \( \{\langle x,b(x)\rangle .\ x\in X\} : X\rightarrow Y \)assumes \( \mathfrak{F} \text{ is a filter on } X \)
shows \( (\text{Net}(\mathfrak{F} )) \text{ is a net on } X \)assumes \( f:A\rightarrow C \)
shows \( domain(f) = A \)assumes \( f:X\rightarrow Y \), \( x\in X \) and \( f = \{\langle x,b(x)\rangle .\ x\in X\} \)
shows \( f(x) = b(x) \) and \( b(x)\in Y \)assumes \( \mathfrak{F} \text{ is a filter on } \bigcup T \) and \( \mathfrak{F} \rightarrow _F x \)
shows \( (\text{Net}(\mathfrak{F} )) \rightarrow _N x \)assumes \( \mathfrak{F} \text{ is a filter on }\bigcup T \) and \( (\text{Net}(\mathfrak{F} )) \rightarrow _N x \)
shows \( \mathfrak{F} \rightarrow _F x \)assumes \( C\subseteq Pow(X) \) and \( C \text{ satisfies the filter base condition } \)
shows \( C \text{ is a base filter } \{A\in Pow(X).\ \exists D\in C.\ D\subseteq A\} \) and \( \bigcup \{A\in Pow(X).\ \exists D\in C.\ D\subseteq A\} = X \)assumes \( C \text{ is a base filter } \mathfrak{F} \)
shows \( (C \text{ satisfies the filter base condition }) \longleftrightarrow (\mathfrak{F} \text{ is a filter on } \bigcup \mathfrak{F} ) \)assumes \( N \text{ is a net on } X \)
shows \( (Filter N.\ .\ X) \text{ is a filter on } X \) and \( \{\{\text{fst}(N)\text{snd}(s).\ s\in \{s\in domain(\text{fst}(N))\times domain(\text{fst}(N)).\ s\in \text{snd}(N) \wedge \text{fst}(s)=t0\}\}.\ t0\in domain(\text{fst}(N))\} \text{ is a base filter } (Filter N.\ .\ X) \)assumes \( \mathfrak{F} \text{ is a filter on } \bigcup T \) and \( C \text{ is a base filter } \mathfrak{F} \) and \( \forall U\in Pow(\bigcup T).\ x\in int(U) \longrightarrow (\exists D\in C.\ D\subseteq U) \) and \( x\in \bigcup T \)
shows \( \mathfrak{F} \rightarrow _F x \)assumes \( \mathfrak{F} \text{ is a filter on } \bigcup T \) and \( C \text{ is a base filter } \mathfrak{F} \) and \( \mathfrak{F} \rightarrow _F x \)
shows \( \forall U\in Pow(\bigcup T).\ x\in int(U) \longrightarrow (\exists D\in C.\ D\subseteq U) \) and \( x\in \bigcup T \)assumes \( N \text{ is a net on } \bigcup T \) and \( (Filter N.\ .\ (\bigcup T)) \rightarrow _F x \)
shows \( N \rightarrow _N x \)assumes \( N \text{ is a net on } \bigcup T \) and \( N \rightarrow _N x \)
shows \( (Filter N.\ .\ (\bigcup T)) \rightarrow _F x \)assumes \( \mathfrak{F} \text{ is a filter on } \bigcup \mathfrak{F} \)
shows \( (\mathfrak{F} \cup \{0\}) \text{ is a topology } \)