This theory continues the series on general topology and covers the definition and basic properties of continuous functions. We also introduce the notion of homeomorphism an prove the pasting lemma.
In this section we define continuous functions and prove that certain conditions are equivalent to a function being continuous.
In standard math we say that a function is continuous with respect to two topologies \(\tau_1 ,\tau_2 \) if the inverse image of sets from topology \(\tau_2\) are in \(\tau_1\). Here we define a predicate that is supposed to reflect that definition, with a difference that we don't require in the definition that \(\tau_1 ,\tau_2 \) are topologies. This means for example that when we define measurable functions, the definition will be the same. The notation \( f^{-1}(A) \) means the inverse image of (a set) \(A\) with respect to (a function) \(f\).
definition
\( \text{IsContinuous}(\tau _1,\tau _2,f) \equiv (\forall U\in \tau _2.\ f^{-1}(U) \in \tau _1) \)
The space of continuous functions mapping \(X=\bigcup \tau_1\) to \(Y=\bigcup \tau_2\) will be denoted \( \text{Cont}(\tau _1,\tau _2) \).
definition
\( \text{Cont}(\tau _1,\tau _2) \equiv \{f\in (\bigcup \tau _1)\rightarrow (\bigcup \tau _2).\ \text{IsContinuous}(\tau _1,\tau _2,f)\} \)
A trivial example of a continuous function - identity is continuous.
lemma id_cont:
shows \( \text{IsContinuous}(\tau ,\tau ,id(\bigcup \tau )) \)proofIdentity is in the space of continuous functions from \(\bigcup \tau\) to itself.
lemma id_cont_sp:
shows \( \{\langle x,x\rangle .\ x\in \bigcup \tau \} \in \text{Cont}(\tau ,\tau ) \)proofA constant function is continuous.
lemma const_cont:
assumes \( T \text{ is a topology } \)
shows \( \text{IsContinuous}(T,\tau , \text{ConstantFunction}(\bigcup T,c)) \)proofIf \(c\in Y = \bigcup S\), then the constant function defined on \(X=\bigcup T \) that is equal to \(c\) is in the the space of continuous functions from \(X\) to \(Y\).
lemma const_cont_sp:
assumes \( T \text{ is a topology } \), \( c\in \bigcup S \)
shows \( \{\langle x,c\rangle .\ x\in \bigcup T\} \in \text{Cont}(T,S) \) using assms, ZF_fun_from_total, const_fun_def_alt, const_cont unfolding Cont_defWe will work with a pair of topological spaces. The following locale sets up our context that consists of two topologies \(\tau_1,\tau_2\) and a continuous function \(f: X_1 \rightarrow X_2\), where \(X_i\) is defined as \(\bigcup\tau_i\) for \(i=1,2\). We also define notation \( cl_1(A) \) and \( cl_2(A) \) for closure of a set \(A\) in topologies \(\tau_1\) and \(\tau_2\), respectively.
locale two_top_spaces0
assumes tau1_is_top: \( \tau _1 \text{ is a topology } \)
assumes tau2_is_top: \( \tau _2 \text{ is a topology } \)
defines \( X_1 \equiv \bigcup \tau _1 \)
defines \( X_2 \equiv \bigcup \tau _2 \)
assumes fmapAssum: \( f: X_1 \rightarrow X_2 \)
defines \( g \text{ is continuous } \equiv \text{IsContinuous}(\tau _1,\tau _2,g) \)
defines \( cl_1(A) \equiv \text{Closure}(A,\tau _1) \)
defines \( cl_2(A) \equiv \text{Closure}(A,\tau _2) \)
First we show that theorems proven in locale topology0 are valid when applied to topologies \(\tau_1\) and \(\tau_2\).
lemma (in two_top_spaces0) topol_cntxs_valid:
shows \( topology0(\tau _1) \) and \( topology0(\tau _2) \) using tau1_is_top, tau2_is_top, topology0_defFor continuous functions the inverse image of a closed set is closed.
lemma (in two_top_spaces0) TopZF_2_1_L1:
assumes A1: \( f \text{ is continuous } \) and A2: \( D \text{ is closed in } \tau _2 \)
shows \( f^{-1}(D) \text{ is closed in } \tau _1 \)proofIf the inverse image of every closed set is closed, then the image of a closure is contained in the closure of the image.
lemma (in two_top_spaces0) Top_ZF_2_1_L2:
assumes A1: \( \forall D.\ ((D \text{ is closed in } \tau _2) \longrightarrow f^{-1}(D) \text{ is closed in } \tau _1) \) and A2: \( A \subseteq X_1 \)
shows \( f(cl_1(A)) \subseteq cl_2(f(A)) \)proofIf \(f\left( \overline{A}\right)\subseteq \overline{f(A)}\) (the image of the closure is contained in the closure of the image), then \(\overline{f^{-1}(B)}\subseteq f^{-1}\left( \overline{B} \right)\) (the inverse image of the closure contains the closure of the inverse image).
lemma (in two_top_spaces0) Top_ZF_2_1_L3:
assumes A1: \( \forall A.\ ( A \subseteq X_1 \longrightarrow f(cl_1(A)) \subseteq cl_2(f(A))) \)
shows \( \forall B.\ ( B \subseteq X_2 \longrightarrow cl_1(f^{-1}(B)) \subseteq f^{-1}(cl_2(B)) ) \)proofIf \(\overline{f^{-1}(B)}\subseteq f^{-1}\left( \overline{B} \right)\) (the inverse image of a closure contains the closure of the inverse image), then the function is continuous. This lemma closes a series of implications in lemmas Top_ZF_2_1_L1, Top_ZF_2_1_L2 and Top_ZF_2_1_L3 showing equivalence of four definitions of continuity.
lemma (in two_top_spaces0) Top_ZF_2_1_L4:
assumes A1: \( \forall B.\ ( B \subseteq X_2 \longrightarrow cl_1(f^{-1}(B)) \subseteq f^{-1}(cl_2(B)) ) \)
shows \( f \text{ is continuous } \)proofFor continuous functions the closure of the inverse image is contained in the inverse image of the closure. This is a shortcut through a series of implications provided by TopZF_2_1_L1, Top_ZF_2_1_L2 and Top_ZF_2_1_L3.
corollary (in two_top_spaces0) im_cl_in_cl_im:
assumes \( f \text{ is continuous } \) and \( B \subseteq X_2 \)
shows \( cl_1(f^{-1}(B)) \subseteq f^{-1}(cl_2(B)) \) using assms, TopZF_2_1_L1, Top_ZF_2_1_L2, Top_ZF_2_1_L3Another condition for continuity: it is sufficient to check if the inverse image of every set in a base is open.
lemma (in two_top_spaces0) Top_ZF_2_1_L5:
assumes A1: \( B \text{ is a base for } \tau _2 \) and A2: \( \forall U\in B.\ f^{-1}(U) \in \tau _1 \)
shows \( f \text{ is continuous } \)proofWe can strenghten the previous lemma: it is sufficient to check if the inverse image of every set in a subbase is open. The proof is rather awkward, as usual when we deal with general intersections. We have to keep track of the case when the collection is empty.
lemma (in two_top_spaces0) Top_ZF_2_1_L6:
assumes A1: \( B \text{ is a subbase for } \tau _2 \) and A2: \( \forall U\in B.\ f^{-1}(U) \in \tau _1 \)
shows \( f \text{ is continuous } \)proofA dual of Top_ZF_2_1_L5: a function that maps base sets to open sets is open.
lemma (in two_top_spaces0) base_image_open:
assumes A1: \( \mathcal{B} \text{ is a base for } \tau _1 \) and A2: \( \forall B\in \mathcal{B} .\ f(B) \in \tau _2 \) and A3: \( U\in \tau _1 \)
shows \( f(U) \in \tau _2 \)proofA composition of two continuous functions is continuous.
lemma comp_cont:
assumes \( \text{IsContinuous}(T,S,f) \) and \( \text{IsContinuous}(S,R,g) \)
shows \( \text{IsContinuous}(T,R,g\circ f) \) using assms, IsContinuous_def, vimage_compA composition of three continuous functions is continuous.
lemma comp_cont3:
assumes \( \text{IsContinuous}(T,S,f) \) and \( \text{IsContinuous}(S,R,g) \) and \( \text{IsContinuous}(R,P,h) \)
shows \( \text{IsContinuous}(T,P,h\circ g\circ f) \) using assms, IsContinuous_def, vimage_compThe graph of a continuous function into a Hausdorff topological space is closed in the product topology. Recall that in ZF a function is the same as its graph.
lemma (in two_top_spaces0) into_T2_graph_closed:
assumes \( f \text{ is continuous } \), \( \tau _2 \text{ is }T_2 \)
shows \( f \text{ is closed in } \text{ProductTopology}(\tau _1,\tau _2) \)proofThis section studies ''homeomorphisms'' - continous bijections whose inverses are also continuous. Notions that are preserved by (commute with) homeomorphisms are called ''topological invariants''.
Homeomorphism is a bijection that preserves open sets.
definition
\( \text{IsAhomeomorphism}(T,S,f) \equiv \) \( f \in \text{bij}(\bigcup T,\bigcup S) \wedge \text{IsContinuous}(T,S,f) \wedge \text{IsContinuous}(S,T,converse(f)) \)
Inverse (converse) of a homeomorphism is a homeomorphism.
lemma homeo_inv:
assumes \( \text{IsAhomeomorphism}(T,S,f) \)
shows \( \text{IsAhomeomorphism}(S,T,converse(f)) \) using assms, IsAhomeomorphism_def, bij_converse_bij, bij_converse_converseHomeomorphisms are open maps.
lemma homeo_open:
assumes \( \text{IsAhomeomorphism}(T,S,f) \) and \( U\in T \)
shows \( f(U) \in S \) using assms, image_converse, IsAhomeomorphism_def, IsContinuous_defA continuous bijection that is an open map is a homeomorphism.
lemma bij_cont_open_homeo:
assumes \( f \in \text{bij}(\bigcup T,\bigcup S) \) and \( \text{IsContinuous}(T,S,f) \) and \( \forall U\in T.\ f(U) \in S \)
shows \( \text{IsAhomeomorphism}(T,S,f) \) using assms, image_converse, IsAhomeomorphism_def, IsContinuous_defA continuous bijection that maps base to open sets is a homeomorphism.
lemma (in two_top_spaces0) bij_base_open_homeo:
assumes A1: \( f \in \text{bij}(X_1,X_2) \) and A2: \( \mathcal{B} \text{ is a base for } \tau _1 \) and A3: \( \mathcal{C} \text{ is a base for } \tau _2 \) and A4: \( \forall U\in \mathcal{C} .\ f^{-1}(U) \in \tau _1 \) and A5: \( \forall V\in \mathcal{B} .\ f(V) \in \tau _2 \)
shows \( \text{IsAhomeomorphism}(\tau _1,\tau _2,f) \) using assms, tau2_is_top, tau1_is_top, bij_converse_bij, bij_is_fun, two_top_spaces0_def, image_converse, Top_ZF_2_1_L5, IsAhomeomorphism_defA bijection that maps base to base is a homeomorphism.
lemma (in two_top_spaces0) bij_base_homeo:
assumes A1: \( f \in \text{bij}(X_1,X_2) \) and A2: \( \mathcal{B} \text{ is a base for } \tau _1 \) and A3: \( \{f(B).\ B\in \mathcal{B} \} \text{ is a base for } \tau _2 \)
shows \( \text{IsAhomeomorphism}(\tau _1,\tau _2,f) \)proofInterior is a topological invariant.
theorem int_top_invariant:
assumes A1: \( A\subseteq \bigcup T \) and A2: \( \text{IsAhomeomorphism}(T,S,f) \)
shows \( f( \text{Interior}(A,T)) = \text{Interior}(f(A),S) \)proofIn this section we consider various ways a topology may be defined on a set that is the range (or the domain) of a function whose domain (or range) is a topological space.
A bijection from a topological space induces a topology on the range.
theorem bij_induced_top:
assumes A1: \( T \text{ is a topology } \) and A2: \( f \in \text{bij}(\bigcup T,Y) \)
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 \( \text{IsAhomeomorphism}(T, \{f(U).\ U\in T\},f) \)proofSuppose we have two topologies \(\tau_1,\tau_2\) on sets \(X_i=\bigcup\tau_i, i=1,2\). Consider some function \(f:A\rightarrow X_2\), where \(A\subseteq X_1\) (we will call such function ''partial''). In such situation we have two natural possibilities for the pairs of topologies with respect to which this function may be continuous. One is obviously the original \(\tau_1,\tau_2\) and in the second one the first element of the pair is the topology relative to the domain of the function: \(\{A\cap U | U \in \tau_1\}\). These two possibilities are not exactly the same and the goal of this section is to explore the differences.
If a function is continuous, then its restriction is continous in relative topology.
lemma (in two_top_spaces0) restr_cont:
assumes A1: \( A \subseteq X_1 \) and A2: \( f \text{ is continuous } \)
shows \( \text{IsContinuous}(\tau _1 \text{ restricted to } A, \tau _2,\text{restrict}(f,A)) \)proofIf a function is continuous, then it is continuous when we restrict the topology on the range to the image of the domain.
lemma (in two_top_spaces0) restr_image_cont:
assumes A1: \( f \text{ is continuous } \)
shows \( \text{IsContinuous}(\tau _1, \tau _2 \text{ restricted to } f(X_1),f) \)proofA combination of restr_cont and restr_image_cont.
lemma (in two_top_spaces0) restr_restr_image_cont:
assumes A1: \( A \subseteq X_1 \) and A2: \( f \text{ is continuous } \) and A3: \( g = \text{restrict}(f,A) \) and A4: \( \tau _3 = \tau _1 \text{ restricted to } A \)
shows \( \text{IsContinuous}(\tau _3, \tau _2 \text{ restricted to } g(A),g) \)proofWe need a context similar to two_top_spaces0 but without the global function \(f:X_1\rightarrow X_2\).
locale two_top_spaces1
assumes tau1_is_top: \( \tau _1 \text{ is a topology } \)
assumes tau2_is_top: \( \tau _2 \text{ is a topology } \)
defines \( X_1 \equiv \bigcup \tau _1 \)
defines \( X_2 \equiv \bigcup \tau _2 \)
If a partial function \(g:X_1\supseteq A\rightarrow X_2\) is continuous with respect to \((\tau_1,\tau_2)\), then \(A\) is open (in \(\tau_1\)) and the function is continuous in the relative topology.
lemma (in two_top_spaces1) partial_fun_cont:
assumes A1: \( g:A\rightarrow X_2 \) and A2: \( \text{IsContinuous}(\tau _1,\tau _2,g) \)
shows \( A \in \tau _1 \) and \( \text{IsContinuous}(\tau _1 \text{ restricted to } A, \tau _2, g) \)proofFor partial function defined on open sets continuity in the whole and relative topologies are the same.
lemma (in two_top_spaces1) part_fun_on_open_cont:
assumes A1: \( g:A\rightarrow X_2 \) and A2: \( A \in \tau _1 \)
shows \( \text{IsContinuous}(\tau _1,\tau _2,g) \longleftrightarrow \) \( \text{IsContinuous}(\tau _1 \text{ restricted to } A, \tau _2, g) \)proofWe start with three topological spaces \((\tau_1,X_1), (\tau_2,X_2)\) and \((\tau_3,X_3)\) and a function \(f:X_1\times X_2\rightarrow X_3\). We will study the properties of \(f\) with respect to the product topology \(\tau_1\times \tau_2\) and \(\tau_3\). This situation is similar as in locale two_top_spaces0 but the first topological space is assumed to be a product of two topological spaces.
First we define a locale with three topological spaces.
locale prod_top_spaces0
assumes tau1_is_top: \( \tau _1 \text{ is a topology } \)
assumes tau2_is_top: \( \tau _2 \text{ is a topology } \)
assumes tau3_is_top: \( \tau _3 \text{ is a topology } \)
defines \( X_1 \equiv \bigcup \tau _1 \)
defines \( X_2 \equiv \bigcup \tau _2 \)
defines \( X_3 \equiv \bigcup \tau _3 \)
defines \( \eta \equiv \text{ProductTopology}(\tau _1,\tau _2) \)
Fixing the first variable in a two-variable continuous function results in a continuous function.
lemma (in prod_top_spaces0) fix_1st_var_cont:
assumes \( f: X_1\times X_2\rightarrow X_3 \) and \( \text{IsContinuous}(\eta ,\tau _3,f) \) and \( x\in X_1 \)
shows \( \text{IsContinuous}(\tau _2,\tau _3, \text{Fix1stVar}(f,x)) \) using assms, fix_1st_var_vimage, IsContinuous_def, tau1_is_top, tau2_is_top, prod_sec_open1Fixing the second variable in a two-variable continuous function results in a continuous function.
lemma (in prod_top_spaces0) fix_2nd_var_cont:
assumes \( f: X_1\times X_2\rightarrow X_3 \) and \( \text{IsContinuous}(\eta ,\tau _3,f) \) and \( y\in X_2 \)
shows \( \text{IsContinuous}(\tau _1,\tau _3, \text{Fix2ndVar}(f,y)) \) using assms, fix_2nd_var_vimage, IsContinuous_def, tau1_is_top, tau2_is_top, prod_sec_open2Having two constinuous mappings we can construct a third one on the cartesian product of the domains.
lemma cart_prod_cont:
assumes A1: \( \tau _1 \text{ is a topology } \), \( \tau _2 \text{ is a topology } \) and A2: \( \eta _1 \text{ is a topology } \), \( \eta _2 \text{ is a topology } \) and A3a: \( f_1:\bigcup \tau _1\rightarrow \bigcup \eta _1 \) and A3b: \( f_2:\bigcup \tau _2\rightarrow \bigcup \eta _2 \) and A4: \( \text{IsContinuous}(\tau _1,\eta _1,f_1) \), \( \text{IsContinuous}(\tau _2,\eta _2,f_2) \) and A5: \( g = \{\langle p,\langle f_1(\text{fst}(p)),f_2(\text{snd}(p))\rangle \rangle .\ p \in \bigcup \tau _1\times \bigcup \tau _2\} \)
shows \( \text{IsContinuous}( \text{ProductTopology}(\tau _1,\tau _2), \text{ProductTopology}(\eta _1,\eta _2),g) \)proofA reformulation of the cart_prod_cont lemma above in slightly different notation.
theorem (in two_top_spaces0) product_cont_functions:
assumes \( f:X_1\rightarrow X_2 \), \( g:\bigcup \tau _3\rightarrow \bigcup \tau _4 \), \( \text{IsContinuous}(\tau _1,\tau _2,f) \), \( \text{IsContinuous}(\tau _3,\tau _4,g) \), \( \tau _4\text{ is a topology } \), \( \tau _3\text{ is a topology } \)
shows \( \text{IsContinuous}( \text{ProductTopology}(\tau _1,\tau _3), \text{ProductTopology}(\tau _2,\tau _4),\{\langle \langle x,y\rangle ,\langle fx,gy\rangle \rangle .\ \langle x,y\rangle \in X_1\times \bigcup \tau _3\}) \)proofA special case of cart_prod_cont when the function acting on the second axis is the identity.
lemma cart_prod_cont1:
assumes A1: \( \tau _1 \text{ is a topology } \) and A1a: \( \tau _2 \text{ is a topology } \) and A2: \( \eta _1 \text{ is a topology } \) and A3: \( f_1:\bigcup \tau _1\rightarrow \bigcup \eta _1 \) and A4: \( \text{IsContinuous}(\tau _1,\eta _1,f_1) \) and A5: \( g = \{\langle p, \langle f_1(\text{fst}(p)),\text{snd}(p)\rangle \rangle .\ p \in \bigcup \tau _1\times \bigcup \tau _2\} \)
shows \( \text{IsContinuous}( \text{ProductTopology}(\tau _1,\tau _2), \text{ProductTopology}(\eta _1,\tau _2),g) \)proofHaving two continuous mappings \(f,g\) we can construct a third one with values in the cartesian product of the codomains of \(f,g\), defined by \(x\mapsto \langle f(x),g(x) \rangle\).
lemma (in prod_top_spaces0) cont_funcs_prod:
assumes \( f:X_1\rightarrow X_2 \), \( g:X_1\rightarrow X_3 \), \( \text{IsContinuous}(\tau _1,\tau _2,f) \), \( \text{IsContinuous}(\tau _1,\tau _3,g) \)
defines \( h \equiv \{\langle x,\langle f(x),g(x)\rangle \rangle .\ x\in X_1\} \)
shows \( \text{IsContinuous}(\tau _1, \text{ProductTopology}(\tau _2,\tau _3),h) \)proofHaving two continuous mappings \(f,g\) we can construct a third one with values in the cartesian product of the codomains of \(f,g\), defined by \(x\mapsto \langle f(x),g(x) \rangle\). This is essentially the same as cont_funcs_prod but formulated in a way that is sometimes easier to apply. Recall that \(\tau_2 \times_t \tau_3\) is a notation for the product topology of \(\tau_1\) and \(\tau_2\).
lemma cont_funcs_prod1:
assumes \( \tau _1 \text{ is a topology } \), \( \tau _2 \text{ is a topology } \), \( \tau _3 \text{ is a topology } \) and \( \{\langle x,p(x)\rangle .\ x\in \bigcup \tau _1\} \in \text{Cont}(\tau _1,\tau _2) \), \( \{\langle x,q(x)\rangle .\ x\in \bigcup \tau _1\} \in \text{Cont}(\tau _1,\tau _3) \)
shows \( \{\langle x,\langle p(x),q(x)\rangle \rangle .\ x\in \bigcup \tau _1\} \in \text{Cont}(\tau _1,\tau _2\times _t\tau _3) \)proofTwo continuous functions into a Hausdorff space are equal on a closed set. Note that in the lemma below \(f\) is assumed to map \(X_1\) to \(X_2\) in the locale, we only need to add a similar assumption for the second function.
lemma (in two_top_spaces0) two_fun_eq_closed:
assumes \( g:X_1\rightarrow X_2 \), \( f \text{ is continuous } \), \( g \text{ is continuous } \), \( \tau _2 \text{ is }T_2 \)
shows \( \{x\in X_1.\ f(x)=g(x)\} \text{ is closed in } \tau _1 \)proofClosure of an image of a singleton by a relation in \(X\times Y\) is contained in the image of this singleton by the closure of the relation (in the product topology). Compare the proof of Metamath's theorem with the same name.
lemma imasncls:
assumes \( T \text{ is a topology } \), \( S \text{ is a topology } \), \( R \subseteq (\bigcup T)\times (\bigcup S) \), \( x\in \bigcup T \)
shows \( \text{Closure}(R\{x\},S) \subseteq \text{Closure}(R,T\times _tS)\{x\} \)proofThe classical pasting lemma states that if \(U_1,U_2\) are both open (or closed) and a function is continuous when restricted to both \(U_1\) and \(U_2\) then it is continuous when restricted to \(U_1 \cup U_2\). In this section we prove a generalization statement stating that the set \(\{ U \in \tau_1 | f|_U\) is continuous \(\}\) is a topology.
A typical statement of the pasting lemma uses the notion of a function restricted to a set being continuous without specifying the topologies with respect to which this continuity holds. In two_top_spaces0 context the notation \( g \text{ is continuous } \) means continuity wth respect to topologies \(\tau_1, \tau_2\). The next lemma is a special case of partial_fun_cont and states that if for some set \(A\subseteq X_1=\bigcup \tau_1\) the function \(f|_A\) is continuous (with respect to \((\tau_1, \tau_2)\)), then \(A\) has to be open. This clears up terminology and indicates why we need to pay attention to the issue of which topologies we talk about when we say that the restricted (to some closed set for example) function is continuos.
lemma (in two_top_spaces0) restriction_continuous1:
assumes A1: \( A \subseteq X_1 \) and A2: \( \text{restrict}(f,A) \text{ is continuous } \)
shows \( A \in \tau _1 \)proofIf a fuction is continuous on each set of a collection of open sets, then it is continuous on the union of them. We could use continuity with respect to the relative topology here, but we know that on open sets this is the same as the original topology.
lemma (in two_top_spaces0) pasting_lemma1:
assumes A1: \( M \subseteq \tau _1 \) and A2: \( \forall U\in M.\ \text{restrict}(f,U) \text{ is continuous } \)
shows \( \text{restrict}(f,\bigcup M) \text{ is continuous } \)proofIf a function is continuous on two sets, then it is continuous on intersection.
lemma (in two_top_spaces0) cont_inter_cont:
assumes A1: \( A \subseteq X_1 \), \( B \subseteq X_1 \) and A2: \( \text{restrict}(f,A) \text{ is continuous } \), \( \text{restrict}(f,B) \text{ is continuous } \)
shows \( \text{restrict}(f,A\cap B) \text{ is continuous } \)proofThe collection of open sets \(U\) such that \(f\) restricted to \(U\) is continuous, is a topology.
theorem (in two_top_spaces0) pasting_theorem:
shows \( \{U \in \tau _1.\ \text{restrict}(f,U) \text{ is continuous }\} \text{ is a topology } \)proof0 is continuous.
corollary (in two_top_spaces0) zero_continuous:
shows \( 0 \text{ is continuous } \)proofassumes \( A\subseteq X \)
shows \( id(X)^{-1}(A) = A \)assumes \( T \text{ is a topology } \)
shows \( (\bigcup T) \in T \)assumes \( c\in A \)
shows \( \text{ConstantFunction}(X,c)^{-1}(A) = X \)assumes \( T \text{ is a topology } \)
shows \( 0 \in T \)assumes \( c\notin A \)
shows \( \text{ConstantFunction}(X,c)^{-1}(A) = \emptyset \)assumes \( \forall x\in X.\ b(x) \in Y \)
shows \( \{\langle x,b(x)\rangle .\ x\in X\} : X\rightarrow Y \)assumes \( T \text{ is a topology } \)
shows \( \text{IsContinuous}(T,\tau , \text{ConstantFunction}(\bigcup T,c)) \)assumes \( f:X\rightarrow Y \)
shows \( f^{-1}(D) \subseteq X \)assumes \( f:X\rightarrow Y \)
shows \( f^{-1}(Y) = X \)assumes \( A\in T \)
shows \( (\bigcup T - A) \text{ is closed in } T \)assumes \( f:X\rightarrow Y \)
shows \( f(B) \subseteq \text{range}(f) \) and \( f(B) \subseteq Y \)assumes \( A \subseteq \bigcup T \)
shows \( A \subseteq cl(A) \)assumes \( f:X\rightarrow Y \) and \( A\subseteq X \)
shows \( A \subseteq f^{-1}(f(A)) \)assumes \( A\subseteq B \)
shows \( f(A)\subseteq f(B) \)assumes \( A \subseteq \bigcup T \)
shows \( cl(A) \text{ is closed in } T \) and \( \bigcup T - cl(A) \in T \)assumes \( B \text{ is closed in } T \), \( A\subseteq B \)
shows \( cl(A) \subseteq B \)assumes \( B \subseteq \bigcup T \) and \( A\subseteq B \)
shows \( cl(A) \subseteq cl(B) \)assumes \( A \subseteq \bigcup T \)
shows \( A \text{ is closed in } T \longleftrightarrow cl(A) = A \)assumes \( f \text{ is continuous } \) and \( D \text{ is closed in } \tau _2 \)
shows \( f^{-1}(D) \text{ is closed in } \tau _1 \)assumes \( \forall D.\ ((D \text{ is closed in } \tau _2) \longrightarrow f^{-1}(D) \text{ is closed in } \tau _1) \) and \( A \subseteq X_1 \)
shows \( f(cl_1(A)) \subseteq cl_2(f(A)) \)assumes \( \forall A.\ ( A \subseteq X_1 \longrightarrow f(cl_1(A)) \subseteq cl_2(f(A))) \)
shows \( \forall B.\ ( B \subseteq X_2 \longrightarrow cl_1(f^{-1}(B)) \subseteq f^{-1}(cl_2(B)) ) \)assumes \( f^{-1}(A) \neq \emptyset \)
shows \( A\neq \emptyset \)assumes \( \bigcap A \neq \emptyset \)
shows \( A\neq \emptyset \)assumes \( \forall V\in B.\ K(V)\in C \) and \( N \in \text{FinPow}(B) \)
shows \( \{K(V).\ V\in N\} \in \text{FinPow}(C) \)assumes \( N\neq 0 \), \( N \in \text{FinPow}(T) \)
shows \( \bigcap N \in T \)assumes \( B \subseteq Pow(Y) \) and \( B\neq \emptyset \) and \( f:X\rightarrow Y \)
shows \( f^{-1}(\bigcap B) = (\bigcap U\in B.\ f^{-1}(U)) \)assumes \( B \text{ is a base for } \tau _2 \) and \( \forall U\in B.\ f^{-1}(U) \in \tau _1 \)
shows \( f \text{ is continuous } \)assumes \( B \text{ is a base for } T \) and \( U\in T \)
shows \( \exists A\in Pow(B).\ U = \bigcup A \)assumes \( B \text{ is a base for } T \)
shows \( \bigcup T = \bigcup B \)assumes \( f:X\rightarrow Y \) and \( \forall A\in M.\ A\subseteq X \)
shows \( f(\bigcup M) = \bigcup \{f(A).\ A\in M\} \)assumes \( f:X\rightarrow Y \)
shows \( f = \{\langle x, f(x)\rangle .\ x \in X\} \)assumes \( f:X\rightarrow Y \)
shows \( f^{-1}(A) = \{x\in X.\ f(x) \in A\} \)assumes \( f:X\rightarrow Y \)
shows \( function(f) \)assumes \( f:X\rightarrow Y \), \( A\cap B = \emptyset \)
shows \( f^{-1}(A)\times B \cap f = \emptyset \)assumes \( T \text{ is a topology } \), \( S \text{ is a topology } \) and \( \forall p\in V.\ \exists U\in T.\ \exists W\in S.\ p\in U\times W \wedge U\times W \subseteq V \)
shows \( V \in \text{ProductTopology}(T,S) \)assumes \( f:X\rightarrow Y \)
shows \( f \subseteq X\times Y \)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 \( a \in \text{bij}(A,B) \)
shows \( converse(converse(a)) = a \)assumes \( f \in \text{inj}(X,Y) \) and \( A\subseteq X \)
shows \( f^{-1}(f(A)) = A \)assumes \( B \text{ is a base for } T \) and \( U \in B \)
shows \( U \in T \)assumes \( \mathcal{B} \text{ is a base for } \tau _1 \) and \( \forall B\in \mathcal{B} .\ f(B) \in \tau _2 \) and \( U\in \tau _1 \)
shows \( f(U) \in \tau _2 \)assumes \( f \in \text{bij}(\bigcup T,\bigcup S) \) and \( \text{IsContinuous}(T,S,f) \) and \( \forall U\in T.\ f(U) \in S \)
shows \( \text{IsAhomeomorphism}(T,S,f) \)assumes \( \text{IsAhomeomorphism}(T,S,f) \) and \( U\in T \)
shows \( f(U) \in S \)assumes \( f \in \text{surj}(X,Y) \) and \( A\subseteq Y \)
shows \( f(f^{-1}(A)) = A \)assumes \( f \in \text{surj}(X,Y) \) and \( B \subseteq Pow(Y) \)
shows \( \{ f(U).\ U \in \{f^{-1}(V).\ V\in B\} \} = B \)assumes \( f \in \text{inj}(X,Y) \) and \( A\subseteq X \), \( B\subseteq X \)
shows \( f(A\cap B) = f(A) \cap f(B) \)assumes \( f:X\rightarrow Y \) and \( A\subseteq X \)
shows \( f(A) = \{f(x).\ x \in A\} \)assumes \( f \in \text{surj}(A,B) \)
shows \( f(A) = B \)assumes \( f:X\rightarrow Y \) and \( B\subseteq X \)
shows \( \text{restrict}(f,B)^{-1}(A) = f^{-1}(A) \cap B \)assumes \( f:X\rightarrow Y \)
shows \( f^{-1}(A \cap f(X)) = f^{-1}(A) \)assumes \( A \subseteq X_1 \) and \( f \text{ is continuous } \)
shows \( \text{IsContinuous}(\tau _1 \text{ restricted to } A, \tau _2,\text{restrict}(f,A)) \)assumes \( f \text{ is continuous } \)
shows \( \text{IsContinuous}(\tau _1, \tau _2 \text{ restricted to } f(X_1),f) \)assumes \( g:A\rightarrow X_2 \) and \( \text{IsContinuous}(\tau _1,\tau _2,g) \)
shows \( A \in \tau _1 \) and \( \text{IsContinuous}(\tau _1 \text{ restricted to } A, \tau _2, g) \)assumes \( f : X\times Y \rightarrow Z \) and \( x\in X \)
shows \( \text{Fix1stVar}(f,x)^{-1}(A) = \{y\in Y.\ \langle x,y\rangle \in f^{-1}(A)\} \)assumes \( T \text{ is a topology } \), \( S \text{ is a topology } \) and \( V \in \text{ProductTopology}(T,S) \) and \( x \in \bigcup T \)
shows \( \{y \in \bigcup S.\ \langle x,y\rangle \in V\} \in S \)assumes \( f : X\times Y \rightarrow Z \) and \( y\in Y \)
shows \( \text{Fix2ndVar}(f,y)^{-1}(A) = \{x\in X.\ \langle x,y\rangle \in f^{-1}(A)\} \)assumes \( T \text{ is a topology } \), \( S \text{ is a topology } \) and \( V \in \text{ProductTopology}(T,S) \) and \( y \in \bigcup S \)
shows \( \{x \in \bigcup T.\ \langle x,y\rangle \in V\} \in T \)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 \( f_1:X_1\rightarrow Y_1 \), \( f_2:X_2\rightarrow Y_2 \) and \( g = \{\langle p,\langle f_1(\text{fst}(p)),f_2(\text{snd}(p))\rangle \rangle .\ p \in X_1\times X_2\} \)
shows \( g^{-1}(A_1\times A_2) = f_1^{-1}(A_1) \times f_2^{-1}(A_2) \)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 \( \tau _1 \text{ is a topology } \), \( \tau _2 \text{ is a topology } \) and \( \eta _1 \text{ is a topology } \), \( \eta _2 \text{ is a topology } \) and \( f_1:\bigcup \tau _1\rightarrow \bigcup \eta _1 \) and \( f_2:\bigcup \tau _2\rightarrow \bigcup \eta _2 \) and \( \text{IsContinuous}(\tau _1,\eta _1,f_1) \), \( \text{IsContinuous}(\tau _2,\eta _2,f_2) \) and \( g = \{\langle p,\langle f_1(\text{fst}(p)),f_2(\text{snd}(p))\rangle \rangle .\ p \in \bigcup \tau _1\times \bigcup \tau _2\} \)
shows \( \text{IsContinuous}( \text{ProductTopology}(\tau _1,\tau _2), \text{ProductTopology}(\eta _1,\eta _2),g) \)assumes \( f:X\rightarrow Y \), \( g:X\rightarrow Z \)
defines \( h \equiv \{\langle x,\langle f(x),g(x)\rangle \rangle .\ x\in X\} \)
shows \( h:X\rightarrow Y\times Z \), \( \forall x\in X.\ h(x) = \langle f(x),g(x)\rangle \), \( h^{-1}(U\times V) = f^{-1}(U) \cap g^{-1}(V) \), \( \{x\in X.\ f(x) = g(x)\} = h^{-1}(\{\langle y,y\rangle .\ y\in Y\}) \)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 \( T \text{ is a topology } \), \( S \text{ is a topology } \)
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 \( \{\langle x,p(x)\rangle .\ x\in X\}: X\rightarrow Y \), \( \{\langle x,q(x)\rangle .\ x\in X\}: X\rightarrow Z \)
defines \( h \equiv \{\langle x,\langle p(x),q(x)\rangle \rangle .\ x\in X\} \)
shows \( h:X\rightarrow Y\times Z \) and \( \forall x\in X.\ h(x) = \langle p(x),q(x)\rangle \)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 \( x\in X \)
shows \( \{\langle x,b(x)\rangle .\ x\in X\}(x)=b(x) \)assumes \( \forall x\in X.\ p(x) = q(x) \)
shows \( \{p(x).\ x\in X\} = \{q(x).\ x\in X\} \) and \( \{\langle x,p(x)\rangle .\ x\in X\} = \{\langle x,q(x)\rangle .\ x\in X\} \)assumes \( f:X\rightarrow Y \), \( g:X\rightarrow Z \)
defines \( h \equiv \{\langle x,\langle f(x),g(x)\rangle \rangle .\ x\in X\} \)
shows \( h:X\rightarrow Y\times Z \), \( \forall x\in X.\ h(x) = \langle f(x),g(x)\rangle \), \( h^{-1}(U\times V) = f^{-1}(U) \cap g^{-1}(V) \), \( \{x\in X.\ f(x) = g(x)\} = h^{-1}(\{\langle y,y\rangle .\ y\in Y\}) \)assumes \( f:X\rightarrow Y \), \( g:X\rightarrow Z \)
defines \( h \equiv \{\langle x,\langle f(x),g(x)\rangle \rangle .\ x\in X\} \)
shows \( h:X\rightarrow Y\times Z \), \( \forall x\in X.\ h(x) = \langle f(x),g(x)\rangle \), \( h^{-1}(U\times V) = f^{-1}(U) \cap g^{-1}(V) \), \( \{x\in X.\ f(x) = g(x)\} = h^{-1}(\{\langle y,y\rangle .\ y\in Y\}) \)assumes \( f:X_1\rightarrow X_2 \), \( g:X_1\rightarrow X_3 \), \( \text{IsContinuous}(\tau _1,\tau _2,f) \), \( \text{IsContinuous}(\tau _1,\tau _3,g) \)
defines \( h \equiv \{\langle x,\langle f(x),g(x)\rangle \rangle .\ x\in X_1\} \)
shows \( \text{IsContinuous}(\tau _1, \text{ProductTopology}(\tau _2,\tau _3),h) \)assumes \( T \text{ is a topology } \)
shows \( T \text{ is }T_2 \longleftrightarrow \{\langle x,x\rangle .\ x\in \bigcup T\} \text{ is closed in } \text{ProductTopology}(T,T) \)assumes \( f:X\rightarrow Y \), \( g:X\rightarrow Z \)
defines \( h \equiv \{\langle x,\langle f(x),g(x)\rangle \rangle .\ x\in X\} \)
shows \( h:X\rightarrow Y\times Z \), \( \forall x\in X.\ h(x) = \langle f(x),g(x)\rangle \), \( h^{-1}(U\times V) = f^{-1}(U) \cap g^{-1}(V) \), \( \{x\in X.\ f(x) = g(x)\} = h^{-1}(\{\langle y,y\rangle .\ y\in Y\}) \)assumes \( T \text{ is a topology } \), \( c\in \bigcup S \)
shows \( \{\langle x,c\rangle .\ x\in \bigcup T\} \in \text{Cont}(T,S) \)assumes \( \tau _1 \text{ is a topology } \), \( \tau _2 \text{ is a topology } \), \( \tau _3 \text{ is a topology } \) and \( \{\langle x,p(x)\rangle .\ x\in \bigcup \tau _1\} \in \text{Cont}(\tau _1,\tau _2) \), \( \{\langle x,q(x)\rangle .\ x\in \bigcup \tau _1\} \in \text{Cont}(\tau _1,\tau _3) \)
shows \( \{\langle x,\langle p(x),q(x)\rangle \rangle .\ x\in \bigcup \tau _1\} \in \text{Cont}(\tau _1,\tau _2\times _t\tau _3) \)assumes \( f \text{ is continuous } \) and \( B \subseteq X_2 \)
shows \( cl_1(f^{-1}(B)) \subseteq f^{-1}(cl_2(B)) \)assumes \( A \subseteq \bigcup T \)
shows \( cl(A) \subseteq \bigcup T \), \( cl(\bigcup T - A) = \bigcup T - int(A) \)assumes \( f:X\rightarrow Y \) and \( A\subseteq X \)
shows \( \text{restrict}(f,A) : A \rightarrow Y \)assumes \( M \subseteq \tau _1 \) and \( \forall U\in M.\ \text{restrict}(f,U) \text{ is continuous } \)
shows \( \text{restrict}(f,\bigcup M) \text{ is continuous } \)assumes \( A \subseteq X_1 \), \( B \subseteq X_1 \) and \( \text{restrict}(f,A) \text{ is continuous } \), \( \text{restrict}(f,B) \text{ is continuous } \)
shows \( \text{restrict}(f,A\cap B) \text{ is continuous } \)