<The previous article in this series | The table of contents of this series |
description/proof of that for net with directed index set eventually in subset of topological space that converges on space, convergence is on closure of subset
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any net with directed index set eventually in any subset of any topological space that converges on the space, the convergence is on the closure of the subset.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(S\): \(\subseteq T\)
\(D\): \(\in \{\text{ the directed sets }\}\)
\(f\): \(: D \to T\), \(\in \{\text{ the nets eventually in } S\}\)
\(t\): \(\in T\)
//
Statements:
\(f\) converges to \(t\)
\(\implies\)
\(t \in \overline{S}\).
//
2: Note
While the proposition that for any net with directed index set on any subset of any topological space that converges on the space, the convergence is on the closure of the subset has been already proved, in fact, the net does not really need to be on the subset but to be eventually in the subset, according to this proposition.
3: Proof
Whole Strategy: Step 1: suppose that \(t \notin \overline{S}\) and find a contradiction.
Step 1:
Let us suppose that \(f\) converges to \(t\).
Let us suppose that \(t \notin \overline{S}\).
By a local characterization of closure: any point on any topological space is on the closure of any subset if and only if its every neighborhood intersects the subset, there would be a neighborhood of \(t\), \(N_t \subseteq T\), such that \(N_t \cap S = \emptyset\).
There would be a \(d \in D\), such that for each \(d' \in D\) such that \(d \le d'\), \(f (d') \in N_t\), because \(f\) converged to \(t\).
Then, for each \(d'' \in D\), there would be a \(d''' \in D\) such that \(d, d'' \le d'''\), and \(f (d''') \in N_t\), which would imply that \(f (d''') \notin S\), which would mean that \(f\) was not eventually in \(S\), a contradiction.
So, \(t \in \overline{S}\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that compact Hausdorff topological space is normal
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any compact Hausdorff topological space is normal.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the compact Hausdorff topological spaces }\}\)
//
Statements:
\(T \in \{\text{ the normal topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: take any closed \(C_1, C_2\) such that \(C_1 \cap C_2 = \emptyset\); Step 2: for each \(c_1 \in C_1\), for each \(c_2 \in C_2\), take an open neighborhood of \(c_1\), \(U_{c_1, c_2}\), and an open neighborhood of \(c_2\), \(U_{c_2}\), such that \(U_{c_1, c_2} \cap U_{c_2} = \emptyset\), take a finite \(\{U_{c_{2, j}}\}\) that covers \(C_2\), and take \(U_{c_1} := \cap U_{c_1, c_{2, j}}\) and \(U_{C_2} := \cup U_{c_{2, j}}\); Step 3: take a finite \(\{U_{c_{1, j}}\}\) that covers \(C_1\) and take \(\cup U_{c_{1, j}}\) and \(\cap U_{C_2, c_{1, j}}\).
Step 1:
Let \(C_1, C_2 \subseteq T\) be any closed subsets such that \(C_1 \cap C_2 = \emptyset\).
Step 2:
Let \(c_1 \in C_1\) be any, which is fixed throughout this Step.
Let \(c_2 \in C_2\) be any.
There are an open neighborhood of \(c_1\), \(U_{c_1, c_2}\), and an open neighborhood of \(c_2\), \(U_{c_2}\), such that \(U_{c_1, c_2} \cap U_{c_2} = \emptyset\), because \(T\) is Hausdorff: \(U_{c_1, c_2}\) depends on \(c_2\) (also \(U_{c_2}\) depends on \(c_1\) but as \(c_1\) is fixed in this Step, the dependence is not explicitly denoted).
\(\{U_{c_2} \vert c_2 \in C_2\}\) is an open cover of \(C_2\).
\(C_2\) is a compact subset of \(T\), by the proposition that any closed subset of any compact topological space is compact.
So, there is a finite subcover of \(\{U_{c_2} \vert c_2 \in C_2\}\), \(\{U_{c_{2, j}} \vert j \in J\}\).
Let us take \(U_{c_1} := \cap_{j \in J} U_{c_1, c_{2, j}}\) and \(U_{C_2} := \cup_{j \in J} U_{c_{2, j}}\).
\(U_{c_1}\) is an open neighborhood of \(c_1\), as a finite intersection.
\(U_{C_2}\) is an open neighborhood of \(C_2\).
\(U_{c_1} \cap U_{C_2} = \emptyset\), because for each \(j \in J\), \(U_{c_1} \subseteq U_{c_1, c_{2, j}}\), so, \(U_{c_1} \cap U_{c_{2, j}} = \emptyset\), and so \(U_{c_1} \cap U_{C_2} = \emptyset\).
Step 3:
By Step 2, for each \(c_1 \in C_1\), there are an open neighborhood of \(c_1\), \(U_{c_1}\), and an open neighborhood of \(U_{C_2, c_1}\), such that \(U_{c_1} \cap U_{C_2, c_1} = \emptyset\): \(U_{C_2, c_1}\) depends on \(c_1\).
\(\{U_{c_1} \vert c_1 \in C_1\}\) is an open cover of \(C_1\).
\(C_1\) is a compact subset of \(T\), as before.
So, there is a finite subcover of \(\{U_{c_1} \vert c_1 \in C_1\}\), \(\{U_{c_{1, j}} \vert j \in J\}\).
Let us take \(U_{C_1} := \cup_{j \in J} U_{c_{1, j}}\) and \(U_{C_2} := \cap_{j \in J} U_{C_2, c_{1, j}}\).
\(U_{C_1}\) is an open neighborhood of \(C_1\).
\(U_{C_2}\) is an open neighborhood of \(C_2\), as a finite intersection of open neighborhoods.
\(U_{C_1} \cap U_{C_2} = \emptyset\), because for each \(j \in J\), \(U_{C_2} \subseteq U_{C_2, c_{1, j}}\), so, \(U_{c_{1, j}} \cap U_{C_2} = \emptyset\), and so, \(U_{C_1} \cap U_{C_2} = \emptyset\).
So, \(T\) is normal.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that continuous bijection from compact topological space onto Hausdorff topological space is homeomorphism
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any continuous bijection from any compact topological space onto any Hausdorff topological space is a homeomorphism.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the compact topological spaces }\}\)
\(T_2\): \(\in \{\text{ the Hausdorff topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous maps }\} \cap \{\text{ the bijections }\}\)
//
Statements:
\(f \in \{\text{ the homeomorphisms }\}\)
//
2: Proof
Whole Strategy: Step 1: apply the proposition that any closed continuous bijection is a homeomorphism.
Step 1:
Let \(C \subseteq T_1\) be any closed subsets.
\(C\) is a compact subset, by the proposition that any closed subset of any compact topological space is compact.
\(f (C)\) is a compact subset, by the proposition that for any continuous map between any topological spaces, the image of any compact subset of the domain is a compact subset of the codomain.
\(f (C)\) is a closed subset, by the proposition that any compact subset of any Hausdorff topological space is closed.
So, \(f\) is a closed map.
So, by the proposition that any closed continuous bijection is a homeomorphism, \(f\) is a homeomorphism.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that closed continuous bijection is homeomorphism
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any closed continuous bijection is a homeomorphism.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in\{\text{ the closed maps }\} \cap \{\text{ the continuous maps }\} \cap \{\text{ the bijections }\}\)
//
Statements:
\(f \in \{\text{ the homeomorphisms }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f^{-1}\) is continuous.
Step 1:
As \(f\) is a bijection, there is the inverse, \(f^{-1}: T_2 \to T_1\).
Let \(U \subseteq T_1\) be any open subset.
\({f^{-1}}^{-1} (U) = f (U) = f (T_1 \setminus C)\) where \(C \subseteq T_1\) is a closed subset.
\(= f (T_1) \setminus f (C)\), by the proposition that for any injective map, the image of any subset minus any subset is the image of the 1st subset minus the image of the 2nd subset, \(= T_2 \setminus f (C)\), because \(f\) is surjective.
\(f (C) \subseteq T_2\) is closed, because \(f\) is closed, and so, \(T_2 \setminus f (C) \subseteq T_2\) is open.
So, \(f^{-1}\) is continuous.
So, \(f\) is a homeomorphism.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that open continuous bijection is homeomorphism
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any open continuous bijection is a homeomorphism.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in\{\text{ the open maps }\} \cap \{\text{ the continuous maps }\} \cap \{\text{ the bijections }\}\)
//
Statements:
\(f \in \{\text{ the homeomorphisms }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f^{-1}\) is continuous.
Step 1:
As \(f\) is a bijection, there is the inverse, \(f^{-1}: T_2 \to T_1\).
Let \(U \subseteq T_1\) be any open subset.
\({f^{-1}}^{-1} (U) = f (U)\), which is open on \(T_2\), because \(f\) is open.
So, \(f^{-1}\) is continuous.
So, \(f\) is a homeomorphism.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for topological space, subspace, and point on subspace, intersection of neighborhoods basis at point on base space and subspace is neighborhoods basis at point on subspace
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any topological space, any subspace, and any point on the subspace, the intersection of any neighborhoods basis at the point on the base space and the subspace is a neighborhoods basis at the point on the subspace.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\in \{\text{ the topological subspaces of } T'\}\)
\(t\): \(\in T\)
\(B'_t\): \(\in \{\text{ the neighborhoods bases at } t \text{ on } T'\}\), \(= \{N'_{t, j} \vert j \in J\}\), where \(J\) is an index set
\(B_t\): \(= \{N'_{t, j} \cap T \vert j \in J\}\)
//
Statements:
\(B_t \in \{\text{ the neighborhoods bases at } t \text{ on } T\}\)
//
2: Proof
Whole Strategy: Step 1: see that each \(N'_{t, j} \cap T\) is a neighborhood of \(t\) on \(T\); Step 2: see that for each neighborhood of \(t\) on \(T\), \(N_t\), there is an \(N'_{t, j} \cap T\) such that \(N'_{t, j} \cap T \subseteq N_t\); Step 3: conclude the proposition.
Step 1:
Each \(N'_{t, j} \cap T \in B_t\) is a neighborhood of \(t\) on \(T\), by the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace.
Step 2:
Let \(N_t \subseteq T\) be any neighborhood of \(t\) on \(T\).
There is an open neighborhood of \(t\) on \(T\), \(U_t \subseteq T\), such that \(U_t \subseteq N_t\).
\(U_t = U'_t \cap T\), where \(U'_t \subseteq T'\) is an open neighborhood of \(t\) on \(T'\), by the definition of subspace topology.
There is an \(N'_{t, j} \in B'_t\) such that \(N'_{t, j} \subseteq U'_t\), by the definition of neighborhoods basis at point.
\(N'_{t, j} \cap T \subseteq U'_t \cap T = U_t \subseteq N_t\).
Step 3:
So, \(B_t\) is a neighborhoods basis at \(t\) on \(T\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that supremum of lower semicontinuous maps is lower semicontinuous
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that the supremum of any set of lower semicontinuous maps is lower semicontinuous.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(\overline{\mathbb{R}}\): \(= \text{ the extended Euclidean topological space }\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{f_j: T_1 \to \overline{\mathbb{R}} \vert j \in J, f_j \in \{\text{ the lower semicontinuous maps }\}\}\):
\(f\): \(: T_1 \to \overline{\mathbb{R}}\), \(= Sup (\{f_j \vert j \in J\})\), which means that for each \(t_1 \in T_1\), \(f (t_1) = Sup (\{f_j (t_1) \vert j \in J\})\)
//
Statements:
\(f \in \{\text{ the lower semicontinuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: for each \(t_1 \in T_1\) and each \(r \lt f (t_1)\), take an open neighborhood of \(t_1\), \(U_{t_1} \subseteq T_1\), such that \(r \lt f_j (U_{t_1})\), and see that \(r \lt f_j (U_{t_1}) \le f (U_{t_1})\).
Step 1:
Let \(t_1 \in T_1\) be any.
Let \(r \lt f (t_1)\) be any.
There is a \(j \in J\) such that \(r \lt f_j (t_1)\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
As \(f_j\) is lower semicontinuous, there is an open neighborhood of \(t_1\), \(U_{t_1} \subseteq T_1\), such that \(r \lt f_j (U_{t_1})\).
Then, \(r \lt f_j (U_{t_1}) \le f (U_{t_1})\), because \(f\) is the supremum.
So, \(f\) is lower semicontinuous.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that restriction of lower semicontinuous map on domain is lower semicontinuous
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any semicontinuous map, the restriction of the map on any domain is lower semicontinuous.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T'_1\): \(\in \{\text{ the topological spaces }\}\)
\(\overline{\mathbb{R}}\): \(= \text{ the extended Euclidean topological space }\)
\(f\): \(: T'_1 \to \overline{\mathbb{R}}\), \(\in \{\text{ the lower semicontinuous maps }\}\)
\(T_1\): \(\subseteq T'_1\), with the subspace topology
\(f \vert_{T_1}\): \(: T_1 \to \overline{\mathbb{R}}\)
//
Statements:
\(f \vert_{T_1} \in \{\text{ the lower semicontinuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: for each \(t_1 \in T_1\) and each \(r \lt f \vert_{T_1} (t_1)\), take an open neighborhood of \(t_1\), \(U'_{t_1} \subseteq T'_1\), such that \(r \lt f (U'_{t_1})\), and take \(U_{t_1} := U'_{t_1} \cap T_1\).
Step 1:
Let \(t_1 \in T_1\) be any.
Let \(r \in \mathbb{R}\) be any such that \(r \lt f \vert_{T_1} (t_1)\).
As \(r \lt f \vert_{T_1} (t_1) = f (t_1)\) and \(f\) is lower semicontinuous, there is an open neighborhood of \(t_1\), \(U'_{t_1} \subseteq T'_1\), such that \(r \lt f (U'_{t_1})\).
Let us take \(U_{t_1} := U'_{t_1} \cap T_1\), which is an open neighborhood of \(t_1\) on \(T_1\), by the definition of subspace topology.
\(f \vert_{T_1} (U_{t_1}) = f (U_{t_1}) \subseteq f (U'_{t_1})\), which implies that \(r \lt f \vert_{T_1} (U_{t_1})\).
So, \(f \vert_{T_1}\) is lower semicontinuous.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for lower semicontinuous map from subspace of product topological space into \(1\)-dimensional extended Euclidean topological space and element of subproduct, cross section of map is lower semicontinuous
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any lower semicontinuous map from any subspace of any product topological space into the \(1\)-dimensional extended Euclidean topological space and any element of any subproduct, the cross section of the map by the element is lower semicontinuous.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(J'\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_{j'} \in \{\text{ the topological spaces }\} \vert j' \in J'\}\):
\(\times_{j' \in J'} T_{j'}\): \(= \text{ the product topological space }\)
\(T\): \(\subseteq \times_{j' \in J'} T_{j'}\), with the subspace topology
\(\overline{\mathbb{R}}\): \(= \text{ the extended Euclidean topological space }\)
\(f\): \(: T \to \overline{\mathbb{R}}\)
\(J\): \(\subset J'\), such that \(J \neq \emptyset\)
\(\times_{j \in J} t_j\): \(\in \times_{j \in J} T_j\)
\(f_{[\times_{j \in J} t_j]}\): \(: T_{[\times_{j \in J} t_j]} \to \overline{\mathbb{R}}\), \(= \text{ the cross section }\)
//
Statements:
\(f \in \{\text{ the lower semicontinuous maps }\}\)
\(\implies\)
\(f_{[\times_{j \in J} t_j]} \in \{\text{ the lower semicontinuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: for each \(\times_{l \in J' \setminus J} t_l \in T_{[\times_{j \in J} t_j]}\) and each \(r \lt f_{[\times_{j \in J} t_j]} (\times_{l \in J' \setminus J} t_l)\), take a \(U_{\times_{j' \in J'} t_{j'}}\) such that \(r \lt f (U_{\times_{j' \in J'} t_{j'}})\), and see that \((U_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\) is an open neighborhood of \(\times_{l \in J' \setminus J} t_l\) and \(r \lt f_{[\times_{j \in J} t_j]} ((U_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]})\).
Step 1:
Let \(\times_{l \in J' \setminus J} t_l \in T_{[\times_{j \in J} t_j]}\) be any.
Let \(r \in \mathbb{R}\) be any such that \(r \lt f_{[\times_{j \in J} t_j]} (\times_{l \in J' \setminus J} t_l)\).
\(f_{[\times_{j \in J} t_j]} (\times_{l \in J' \setminus J} t_l) = f (\times_{j' \in J'} t_{j'})\).
As \(r \lt f (\times_{j' \in J'} t_{j'})\) and \(f\) is lower semicontinuous, there is an open neighborhood of \(\times_{j' \in J'} t_{j'}\), \(U_{\times_{j' \in J'} t_{j'}} \subseteq T\), such that \(r \lt f (U_{\times_{j' \in J'} t_{j'}})\).
\((U_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\), the cross section of \(U_{\times_{j' \in J'} t_{j'}}\) by \(\times_{j \in J} t_j\), is an open neighborhood of \(\times_{l \in J' \setminus J} t_l\) on \(T_{[\times_{j \in J} t_j]}\), because \(\times_{j' \in J'} t_{j'} \in U_{\times_{j' \in J'} t_{j'}}\) and \((U_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]} \subseteq T_{[\times_{j \in J} t_j]}\) is open, by the proposition that for any product topological space, any subspace, any open subset of the subspace, and any element of any subproduct, the cross section of the open subset by the element is open on the cross section of the subspace by the element.
Then, for each \(\times_{l \in J' \setminus J} u_l \in (U_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\), taking for each \(j \in J\), \(u_j := t_j\), \(\times_{j' \in J'} u_l \in U_{\times_{j' \in J'} t_{j'}}\), so, \(f_{[\times_{j \in J} t_j]} (\times_{l \in J' \setminus J} u_l) = f (\times_{j' \in J'} u_l) \in f (U_{\times_{j' \in J'} t_{j'}})\), so, \(r \lt f_{[\times_{j \in J} t_j]} (\times_{l \in J' \setminus J} u_l)\), so, \(r \lt f_{[\times_{j \in J} t_j]} ((U_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]})\).
So, \(f_{[\times_{j \in J} t_j]}\) is lower semicontinuous.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for product topological space, subspace, open subset of subspace, and element of subproduct, cross section of open subset is open on cross section of subspace
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any product topological space, any subspace, any open subset of the subspace, and any element of any subproduct, the cross section of the open subset by the element is open on the cross section of the subspace by the element.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(J'\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_{j'} \in \{\text{ the topological spaces }\} \vert j' \in J'\}\):
\(\times_{j' \in J'} T_{j'}\): \(= \text{ the product topological space }\)
\(T\): \(\subseteq \times_{j' \in J'} T_{j'}\), with the subspace topology
\(U\): \(\in \{\text{ the open subsets of } T\}\)
\(J\): \(\subset J'\), such that \(J \neq \emptyset\)
\(\times_{j \in J} t_j\): \(\in \times_{j \in J} T_j\)
\(U_{[\times_{j \in J} t_j]}\): \(= \text{ the cross section }\)
\(T_{[\times_{j \in J} t_j]}\): \(= \text{ the cross section } \subseteq \times_{l \in J' \setminus J} T_l\) with the subspace topology
//
Statements:
\(U_{[\times_{j \in J} t_j]} \in \{\text{ the open subsets of } T_{[\times_{j \in J} t_j]}\}\)
//
2: Note
When \(T = \times_{j' \in J'} T_{j'}\), \(U\) is open on \(\times_{j' \in J'} T_{j'}\), and as \(T_{[\times_{j \in J} t_j]} = \times_{l \in J' \setminus J} T_l\), \(U_{[\times_{j \in J} t_j]}\) is open on \(\times_{l \in J' \setminus J} T_l\).
3: Proof
Whole Strategy: Step 1: for each \(\times_{l \in J' \setminus J} t_l \in U_{[\times_{j \in J} t_j]}\), \(\times_{j' \in J'} t_{j'} \in U\), and take an open neighborhood of \(\times_{j' \in J'} t_{j'}\), \(U_{\times_{j' \in J'} t_{j'}} = U'_{\times_{j' \in J'} t_{j'}} \cap T \subseteq U\), and see that \((U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]} \cap T_{[\times_{j \in J} t_j]} \subseteq U_{[\times_{j \in J} t_j]}\); Step 2: see that \((U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\) is an open neighborhood of \(\times_{l \in J' \setminus J} t_l\) on \(\times_{l \in J' \setminus J} T_l\).
Step 1:
Let \(\times_{l \in J' \setminus J} t_l \in U_{[\times_{j \in J} t_j]}\) be any.
\(\times_{j' \in J'} t_{j'} \in U\).
There is an open neighborhood of \(\times_{j' \in J'} t_{j'}\), \(U_{\times_{j' \in J'} t_{j'}} \subseteq T\), such that \(U_{\times_{j' \in J'} t_{j'}} \subseteq U\), by the local criterion for openness.
\(U_{\times_{j' \in J'} t_{j'}} = U'_{\times_{j' \in J'} t_{j'}} \cap T\) where \(U'_{\times_{j' \in J'} t_{j'}} \subseteq \times_{j' \in J'} T_{j'}\) is an open neighborhood of \(\times_{j' \in J'} t_{j'}\) on \(\times_{j' \in J'} T_{j'}\), by the definition of subspace topology.
\((U'_{\times_{j' \in J'} t_{j'}} \cap T)_{[\times_{j \in J} t_j]} \subseteq U_{[\times_{j \in J} t_j]}\), by the proposition that for any product set, any subset, any subset that contains the 1st subset, and any element of any subproduct, the cross section of the 1st subset by the element is contained in the cross section of the 2nd subset by the element.
But \((U'_{\times_{j' \in J'} t_{j'}} \cap T)_{[\times_{j \in J} t_j]} = (U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]} \cap T_{[\times_{j \in J} t_j]}\), by the proposition that for any product set, any subsets, and any element of any subproduct, the cross section of the intersection of the subsets by the element is the intersection of the cross sections of the subsets by the element.
So, \((U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]} \cap T_{[\times_{j \in J} t_j]} \subseteq U_{[\times_{j \in J} t_j]}\).
Step 2:
Let us see that \((U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\) is an open neighborhood of \(\times_{l \in J' \setminus J} t_l\) on \(\times_{l \in J' \setminus J} T_l\).
\(\times_{l \in J' \setminus J} t_l \in (U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\), because \(\times_{j' \in J'} t_{j'} \in U'_{\times_{j' \in J'} t_{j'}}\).
Let \(\times_{l \in J' \setminus J} u_l \in (U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\) be any.
Taking for each \(j \in J\), \(u_j := t_j\), \(\times_{j' \in J'} u_{j'} \in U'_{\times_{j' \in J'} t_{j'}}\).
By Note for the definition of product topology, there is a \(\times_{j' \in J'} U'_{u_{j'}} \subseteq U'_{\times_{j' \in J'} t_{j'}}\), where \(U'_{u_{j'}} \subseteq T_{j'}\) is an open neighborhood of \(u_{j'}\) such that only finite of \(U'_{u_{j'}}\) s are not \(T_{j'}\) s.
\(\times_{l \in J' \setminus J} U'_{u_l} \subseteq \times_{l \in J' \setminus J} T_l\) is an open neighborhood of \(\times_{l \in J' \setminus J} u_l\), by Note for the definition of product topology: only finite of \(U'_{u_l}\) s are not \(T_l\) s.
Let us see that \(\times_{l \in J' \setminus J} U'_{u_l} \subseteq (U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\).
Let \(\times_{l \in J' \setminus J} v_l \in \times_{l \in J' \setminus J} U'_{u_l}\) be any.
Taking for each \(j \in J\), \(v_j := t_j\), \(\times_{j' \in J'} v_{j'} \in \times_{j' \in J'} U'_{u_{j'}} \subseteq U'_{\times_{j' \in J'} t_{j'}}\).
That means that \(\times_{l \in J' \setminus J} v_l \in (U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\).
So, \(\times_{l \in J' \setminus J} U'_{u_l} \subseteq (U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\).
By the local criterion for openness, \((U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]}\) is open on \(\times_{l \in J' \setminus J} T_l\).
So, \((U'_{\times_{j' \in J'} t_{j'}})_{[\times_{j \in J} t_j]} \cap T_{[\times_{j \in J} t_j]}\) is open on \(T_{[\times_{j \in J} t_j]}\), by the definition of subspace topology.
By the local criterion for openness, \(U_{[\times_{j \in J} t_j]}\) is open on \(T_{[\times_{j \in J} t_j]}\).
References
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description/proof of that for product set, subset, and element of subproduct, cross section of subset is contained in projection of subset
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any product set, any subset, and any element of any subproduct, the cross section of the subset by the element is contained in the projection of the subset onto the complemental subproduct.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(J'\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_{j'} \in \{\text{ the sets }\} \vert j' \in J'\}\):
\(\times_{j' \in J'} S_{j'}\): \(= \text{ the product set }\)
\(Q\): \(\subseteq \times_{j' \in J'} S_{j'}\)
\(J\): \(\subset J'\), such that \(J \neq \emptyset\)
\(\times_{j \in J} S_j\): \(= \text{ the product set }\)
\(\times_{j \in J} s_j\): \(\in \times_{j \in J} S_j\)
\(\pi^{J' \setminus J}\): \(: \times_{j' \in J'} S_{j'} \to \times_{l \in J' \setminus J} S_l\), \(= \text{ the projection }\)
//
Statements:
\(Q_{[\times_{j \in J} s_j]} \subseteq \pi^{J' \setminus J} (Q)\)
//
2: Proof
Whole Strategy: Step 1: see that for each \(\times_{l \in J' \setminus J} s_l \in Q_{[\times_{j \in J} s_j]}\), \(\times_{l \in J' \setminus J} s_l \in \pi^{J' \setminus J} (Q)\).
Step 1:
Let \(\times_{l \in J' \setminus J} s_l \in Q_{[\times_{j \in J} s_j]}\) be any.
\(\times_{j' \in J'} s_{j'} \in Q\).
\(\pi^{J' \setminus J} (\times_{j' \in J'} s_{j'}) = \times_{l \in J' \setminus J} s_l\), which means that \(\times_{l \in J' \setminus J} s_l \in \pi^{J' \setminus J} (Q)\).
So, \(Q_{[\times_{j \in J} s_j]} \subseteq \pi^{J' \setminus J} (Q)\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for product set, \(2\) subsets, and element of subproduct, cross section of 1st subset minus 2nd subset is cross section of 1st subset minus cross section of 2nd subset
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any product set, any \(2\) subsets, and any element of any subproduct, the cross section of the 1st subset minus the 2nd subset by the element is the cross section of the 1st subset by the element minus the cross section of the 2nd subset by the element.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(J'\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_{j'} \in \{\text{ the sets }\} \vert j' \in J'\}\):
\(\times_{j' \in J'} S_{j'}\): \(= \text{ the product set }\)
\(Q_1\): \(\subseteq \times_{j' \in J'} S_{j'}\)
\(Q_2\): \(\subseteq \times_{j' \in J'} S_{j'}\)
\(J\): \(\subset J'\), such that \(J \neq \emptyset\)
\(\times_{j \in J} s_j\): \(\in \times_{j \in J} S_j\)
//
Statements:
\((Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]} = (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\)
//
2: Proof
Whole Strategy: Step 1: see that \((Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]} \subseteq (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\); Step 2: see that \((Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]} \subseteq (Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]}\); Step 3: conclude the proposition.
Step 1:
Let \(\times_{l \in J' \setminus J} s_l \in (Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]}\) be any.
\(\times_{j' \in J'} s_{j'} \in Q_1 \setminus Q_2\).
\(\times_{j' \in J'} s_{j'} \in Q_1\) and \(\times_{j' \in J'} s_{j'} \notin Q_2\).
So, \(\times_{l \in J' \setminus J} s_l \in (Q_1)_{[\times_{j \in J} s_j]}\) and \(\times_{l \in J' \setminus J} s_l \notin (Q_2)_{[\times_{j \in J} s_j]}\).
So, \(\times_{l \in J' \setminus J} s_l \in (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\).
So, \((Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]} \subseteq (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\).
Step 2:
Let \(\times_{l \in J' \setminus J} s_l \in (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\) be any.
\(\times_{l \in J' \setminus J} s_l \in (Q_1)_{[\times_{j \in J} s_j]}\) and \(\times_{l \in J' \setminus J} s_l \notin (Q_2)_{[\times_{j \in J} s_j]}\).
\(\times_{j' \in J'} s_{j'} \in Q_1\) and \(\times_{j' \in J'} s_{j'} \notin Q_2\).
So, \(\times_{j' \in J'} s_{j'} \in Q_1 \setminus Q_2\).
So, \(\times_{l \in J' \setminus J} s_l \in (Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]}\).
So, \((Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]} \subseteq (Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]}\).
Step 3:
So, \((Q_1 \setminus Q_2)_{[\times_{j \in J} s_j]} = (Q_1)_{[\times_{j \in J} s_j]} \setminus (Q_2)_{[\times_{j \in J} s_j]}\).
References
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