Topology_ZF_1 theory describes how we can define a topology on a product of two topological spaces. One way to generalize that is to construct topology for a cartesian product of \(n\) topological spaces. The cartesian product approach is somewhat inconvenient though. Another way to approach product topology on \(X^n\) is to model cartesian product as sets of sequences (of length \(n\)) of elements of \(X\). This means that having a topology on \(X\) we want to define a toplogy on the space \(n\rightarrow X\), where \(n\) is a natural number (recall that \(n = \{ 0,1,...,n-1\}\) in ZF). However, this in turn can be done more generally by defining a topology on any function space \(I\rightarrow X\), where \(I\) is any set of indices. This is what we do in this theory.
In this section we define the base of the product topology.
Suppose \(\mathcal{X} = I\rightarrow \bigcup T\) is a space of functions from some index set \(I\) to the carrier of a topology \(T\). Then take a finite collection of open sets \(W:N\rightarrow T\) indexed by \(N\subseteq I\). We can define a subset of \(\mathcal{X}\) that models the cartesian product of \(W\).
definition
\( \text{FinProd}(\mathcal{X} ,W) \equiv \{x\in \mathcal{X} .\ \forall i\in domain(W).\ x(i) \in W(i)\} \)
Now we define the base of the product topology as the collection of all finite products (in the sense defined above) of open sets.
definition
\( \text{ProductTopBase}(I,T) \equiv \bigcup N\in \text{FinPow}(I).\ \{ \text{FinProd}(I\rightarrow \bigcup T,W).\ W\in N\rightarrow T\} \)
Finally, we define the product topology on sequences. We use the ''Seq'' prefix although the definition is good for any index sets, not only natural numbers.
definition
\( \text{SeqProductTopology}(I,T) \equiv \{\bigcup B.\ B \in Pow( \text{ProductTopBase}(I,T))\} \)
Product topology base is closed with respect to intersections.
lemma prod_top_base_inter:
assumes A1: \( T \text{ is a topology } \) and A2: \( U \in \text{ProductTopBase}(I,T) \), \( V \in \text{ProductTopBase}(I,T) \)
shows \( U\cap V \in \text{ProductTopBase}(I,T) \)proofIn the next theorem we show the collection of sets defined above as \( \text{ProductTopBase}(\mathcal{X} ,T) \) satisfies the base condition. This is a condition, defined in Topology_ZF_1 that allows to claim that this collection is a base for some topology.
theorem prod_top_base_is_base:
assumes \( T \text{ is a topology } \)
shows \( \text{ProductTopBase}(I,T) \text{ satisfies the base condition } \) using assms, prod_top_base_inter, inter_closed_baseThe (sequence) product topology is indeed a topology on the space of sequences. In the proof we are using the fact that \((\emptyset\rightarrow X) = \{\emptyset\}\).
theorem seq_prod_top_is_top:
assumes \( T \text{ is a topology } \)
shows \( \text{SeqProductTopology}(I,T) \text{ is a topology } \) and \( \text{ProductTopBase}(I,T) \text{ is a base for } \text{SeqProductTopology}(I,T) \) and \( \bigcup \text{SeqProductTopology}(I,T) = (I\rightarrow \bigcup T) \)proofAs a special case of the space of functions \(I\rightarrow X\) we can consider space of lists of elements of \(X\), i.e. space \(n\rightarrow X\), where \(n\) is a natural number (recall that in ZF set theory \(n=\{0,1,...,n-1\}\)). Such spaces model finite cartesian products \(X^n\) but are easier to deal with in formalized way (than the said products). This section discusses natural topology defined on \(n\rightarrow X\) where \(X\) is a topological space.
When the index set is finite, the definition of \( \text{ProductTopBase}(I,T) \) can be simplifed.
lemma fin_prod_def_nat:
assumes A1: \( n\in nat \) and A2: \( T \text{ is a topology } \)
shows \( \text{ProductTopBase}(n,T) = \{ \text{FinProd}(n\rightarrow \bigcup T,W).\ W\in n\rightarrow T\} \)proofA technical lemma providing a formula for finite product on one topological space.
lemma single_top_prod:
assumes A1: \( W:1\rightarrow \tau \)
shows \( \text{FinProd}(1\rightarrow \bigcup \tau ,W) = \{ \{\langle 0,y\rangle \}.\ y \in W(0)\} \)proofIntuitively, the topological space of singleton lists valued in \(X\) is the same as \(X\). However, each element of this space is a list of length one, i.e a set consisting of a pair \(\langle 0, x\rangle\) where \(x\) is an element of \(X\). The next lemma provides a formula for the product topology in the corner case when we have only one factor and shows that the product topology of one space is essentially the same as the space.
lemma singleton_prod_top:
assumes A1: \( \tau \text{ is a topology } \)
shows \( \text{SeqProductTopology}(1,\tau ) = \{ \{ \{\langle 0,y\rangle \}.\ y\in U \}.\ U\in \tau \} \) and \( \text{IsAhomeomorphism}(\tau , \text{SeqProductTopology}(1,\tau ),\{\langle y,\{\langle 0,y\rangle \}\rangle .\ y \in \bigcup \tau \}) \)proofA special corner case of finite_top_prod_homeo: a space \(X\) is homeomorphic to the space of one element lists of \(X\).
theorem singleton_prod_top1:
assumes A1: \( \tau \text{ is a topology } \)
shows \( \text{IsAhomeomorphism}( \text{SeqProductTopology}(1,\tau ),\tau ,\{\langle x,x(0)\rangle .\ x\in 1\rightarrow \bigcup \tau \}) \)proofA technical lemma describing the carrier of a (cartesian) product topology of the (sequence) product topology of \(n\) copies of topology \(\tau\) and another copy of \(\tau\).
lemma finite_prod_top:
assumes \( \tau \text{ is a topology } \) and \( T = \text{SeqProductTopology}(n,\tau ) \)
shows \( (\bigcup \text{ProductTopology}(T,\tau )) = (n\rightarrow \bigcup \tau )\times \bigcup \tau \) using assms, Top_1_4_T1, seq_prod_top_is_topIf \(U\) is a set from the base of \(X^n\) and \(V\) is open in \(X\), then \(U\times V\) is in the base of \(X^{n+1}\). The next lemma is an analogue of this fact for the function space approach.
lemma finite_prod_succ_base:
assumes A1: \( \tau \text{ is a topology } \) and A2: \( n \in nat \) and A3: \( U \in \text{ProductTopBase}(n,\tau ) \) and A4: \( V\in \tau \)
shows \( \{x \in succ(n)\rightarrow \bigcup \tau .\ \text{Init}(x) \in U \wedge x(n) \in V\} \in \text{ProductTopBase}(succ(n),\tau ) \)proofIf \(U\) is open in \(X^n\) and \(V\) is open in \(X\), then \(U\times V\) is open in \(X^{n+1}\). The next lemma is an analogue of this fact for the function space approach.
lemma finite_prod_succ:
assumes A1: \( \tau \text{ is a topology } \) and A2: \( n \in nat \) and A3: \( U \in \text{SeqProductTopology}(n,\tau ) \) and A4: \( V\in \tau \)
shows \( \{x \in succ(n)\rightarrow \bigcup \tau .\ \text{Init}(x) \in U \wedge x(n) \in V\} \in \text{SeqProductTopology}(succ(n),\tau ) \)proofIn the Topology_ZF_2 theory we define product topology of two topological spaces. The next lemma explains in what sense the topology on finite lists of length \(n\) of elements of topological space \(X\) can be thought as a model of the product topology on the cartesian product of \(n\) copies of that space. Namely, we show that the space of lists of length \(n+1\) of elements of \(X\) is homeomorphic to the product topology (as defined in Topology_ZF_2) of two spaces: the space of lists of length \(n\) and \(X\). Recall that if \(\mathcal{B}\) is a base (i.e. satisfies the base condition), then the collection \(\{\bigcup B| B \in Pow(\mathcal{B})\}\) is a topology (generated by \(\mathcal{B}\)).
theorem finite_top_prod_homeo:
assumes A1: \( \tau \text{ is a topology } \) and A2: \( n \in nat \) and A3: \( f = \{\langle x,\langle \text{Init}(x),x(n)\rangle \rangle .\ x \in succ(n)\rightarrow \bigcup \tau \} \) and A4: \( T = \text{SeqProductTopology}(n,\tau ) \) and A5: \( S = \text{SeqProductTopology}(succ(n),\tau ) \)
shows \( \text{IsAhomeomorphism}(S, \text{ProductTopology}(T,\tau ),f) \)proofassumes \( \forall x\in A\cap B.\ h(x) \in Y \), \( \forall x\in A\setminus B.\ f(x) \in Y \), \( \forall x\in B\setminus A.\ g(x) \in Y \)
shows \( \{\langle x,\text{if }x\in A\setminus B\text{ then }f(x)\text{ else }\text{if }x\in B\setminus A\text{ then }g(x)\text{ else }h(x)\rangle .\ x \in A\cup B\}: A\cup B \rightarrow Y \)assumes \( A \in \text{FinPow}(X) \) and \( B \in \text{FinPow}(X) \)
shows \( A \cup B \in \text{FinPow}(X) \)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 \( T \text{ is a topology } \) and \( U \in \text{ProductTopBase}(I,T) \), \( V \in \text{ProductTopBase}(I,T) \)
shows \( U\cap V \in \text{ProductTopBase}(I,T) \)assumes \( \forall U\in B.\ (\forall V\in B.\ U\cap V \in B) \)
shows \( B \text{ satisfies the base condition } \)assumes \( T \text{ is a topology } \)
shows \( \text{ProductTopBase}(I,T) \text{ satisfies the base condition } \)assumes \( B \text{ satisfies the base condition } \) and \( T = \{\bigcup A.\ A\in Pow(B)\} \)
shows \( T \text{ is a topology } \) and \( B \text{ is a base for } T \)assumes \( B \text{ is a base for } T \)
shows \( \bigcup T = \bigcup B \)assumes \( n \in nat \)
shows \( n \in \text{FinPow}(nat) \)assumes \( A \in \text{FinPow}(X) \)
shows \( A \in \text{FinPow}(A) \)assumes \( \forall x\in X.\ b(x) \in Y \)
shows \( \{\langle x,b(x)\rangle .\ x\in X\} : X\rightarrow Y \)assumes \( f : \{a\}\rightarrow X \)
shows \( f = \{\langle a, f(a)\rangle \} \)assumes \( y \in Y \)
shows \( \{\langle x,y\rangle \} : \{x\} \rightarrow Y \)assumes \( T \text{ is a topology } \) and \( f \in \text{bij}(\bigcup T,Y) \)
shows \( \{f(U).\ U\in T\} \text{ is a topology } \) and \( \{ \{f(x).\ x\in U\}.\ U\in T\} \text{ is a topology } \) and \( (\bigcup \{f(U).\ U\in T\}) = Y \) and \( \text{IsAhomeomorphism}(T, \{f(U).\ U\in T\},f) \)assumes \( f:X\rightarrow Y \) and \( A\subseteq X \)
shows \( f(A) = \{f(x).\ x \in A\} \)assumes \( n\in nat \) and \( T \text{ is a topology } \)
shows \( \text{ProductTopBase}(n,T) = \{ \text{FinProd}(n\rightarrow \bigcup T,W).\ W\in n\rightarrow T\} \)assumes \( W:1\rightarrow \tau \)
shows \( \text{FinProd}(1\rightarrow \bigcup \tau ,W) = \{ \{\langle 0,y\rangle \}.\ y \in W(0)\} \)assumes \( T \text{ is a topology } \)
shows \( \text{SeqProductTopology}(I,T) \text{ is a topology } \) and \( \text{ProductTopBase}(I,T) \text{ is a base for } \text{SeqProductTopology}(I,T) \) and \( \bigcup \text{SeqProductTopology}(I,T) = (I\rightarrow \bigcup T) \)assumes \( B \text{ is a topology } \) and \( B \text{ is a base for } T \)
shows \( B=T \)assumes \( \tau \text{ is a topology } \)
shows \( \text{SeqProductTopology}(1,\tau ) = \{ \{ \{\langle 0,y\rangle \}.\ y\in U \}.\ U\in \tau \} \) and \( \text{IsAhomeomorphism}(\tau , \text{SeqProductTopology}(1,\tau ),\{\langle y,\{\langle 0,y\rangle \}\rangle .\ y \in \bigcup \tau \}) \)assumes \( \text{IsAhomeomorphism}(T,S,f) \)
shows \( \text{IsAhomeomorphism}(S,T,converse(f)) \)assumes \( T \text{ is a topology } \), \( S \text{ is a topology } \)
shows \( (T\times _tS) \text{ is a topology } \), \( \text{ProductCollection}(T,S) \text{ is a base for } (T\times _tS) \), \( \bigcup (T\times _tS) = \bigcup T \times \bigcup S \)assumes \( a: n \rightarrow X \) and \( x\in X \) and \( b = \text{Append}(a,x) \)
shows \( b : succ(n) \rightarrow X \), \( \forall k\in n.\ b(k) = a(k) \), \( b(n) = x \)assumes \( n \in nat \) and \( a: succ(n) \rightarrow X \)
shows \( \text{Init}(a) : n \rightarrow X \), \( \forall k\in n.\ \text{Init}(a)(k) = a(k) \), \( a = \text{Append}( \text{Init}(a), a(n)) \)assumes \( B \text{ is a base for } T \) and \( U\in T \)
shows \( \exists A\in Pow(B).\ U = \bigcup A \)assumes \( \tau \text{ is a topology } \) and \( n \in nat \) and \( U \in \text{ProductTopBase}(n,\tau ) \) and \( V\in \tau \)
shows \( \{x \in succ(n)\rightarrow \bigcup \tau .\ \text{Init}(x) \in U \wedge x(n) \in V\} \in \text{ProductTopBase}(succ(n),\tau ) \)assumes \( B \text{ is a base for } T \) and \( U \in B \)
shows \( U \in T \)assumes \( T \text{ is a topology } \) and \( \forall i\in I.\ P(i) \in T \)
shows \( (\bigcup i\in I.\ P(i)) \in T \)assumes \( n \in nat \)
shows \( \{\langle x,\langle \text{Init}(x),x(n)\rangle \rangle .\ x \in succ(n)\rightarrow X\} \in \text{bij}(succ(n)\rightarrow X,(n\rightarrow X)\times X) \)assumes \( f:X\rightarrow Y \)
shows \( f^{-1}(A) = \{x\in X.\ f(x) \in A\} \)assumes \( \tau \text{ is a topology } \) and \( n \in nat \) and \( U \in \text{SeqProductTopology}(n,\tau ) \) and \( V\in \tau \)
shows \( \{x \in succ(n)\rightarrow \bigcup \tau .\ \text{Init}(x) \in U \wedge x(n) \in V\} \in \text{SeqProductTopology}(succ(n),\tau ) \)assumes \( T \text{ is a topology } \), \( S \text{ is a topology } \) and \( U\in T \), \( V\in S \)
shows \( U\times V \in \text{ProductTopology}(T,S) \)assumes \( f:X\rightarrow Y \) and \( f = \{\langle x,b(x)\rangle .\ x\in X\} \)
shows \( \forall x\in X.\ f(x) = b(x) \)assumes \( n \in nat \) and \( a:n\rightarrow X \) and \( x \in X \)
shows \( \text{Init}( \text{Append}(a,x)) = a \)assumes \( f \in \text{bij}(X_1,X_2) \) and \( \mathcal{B} \text{ is a base for } \tau _1 \) and \( \mathcal{C} \text{ is a base for } \tau _2 \) and \( \forall U\in \mathcal{C} .\ f^{-1}(U) \in \tau _1 \) and \( \forall V\in \mathcal{B} .\ f(V) \in \tau _2 \)
shows \( \text{IsAhomeomorphism}(\tau _1,\tau _2,f) \)