<The previous article in this series | The table of contents of this series |
description/proof of that for locally path-connected topological space, path-connected component is connected component
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 locally path-connected topological space, any path-connected component is a connected component.
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:
\(T\): \(\in \{\text{ the locally path-connected topological spaces }\}\)
\(C\): \(\in \{\text{ the path-connected components of } T\}\)
\(C'\): \(\in \{\text{ the connected components of } T\}\), such that \(c \in C'\) for any fixed \(c \in C\)
//
Statements:
\(C = C'\)
//
2: Note
\(C'\) is uniquely determined, by this proposition, because for any other \(\widetilde{C'}\) taken based on \(\widetilde{c} \in C\), \(C = \widetilde{C'}\) means that \(c \in \widetilde{C'}\), so, \(\widetilde{C'}\) is the connected component such that \(c \in \widetilde{C'}\), which is \(C'\).
3: Proof
Whole Strategy: Step 1: see that \(C \subseteq C'\); Step 2: suppose that \(C \subset C'\), and find a contradiction.
Step 1:
\(C \subseteq C'\), by the proposition that for any topological space, any path-connected component is contained in the corresponding connected component.
Step 2:
Let us suppose that \(C \subset C'\).
For each \(c' \in C'\), there would be the path-connected component, \(C_{c'} \subseteq T\), such that \(c' \in C_{c'}\).
\(C_{c'} \subseteq C'\), by the proposition that for any topological space, any path-connected component is contained in the corresponding connected component.
\(C_{c'}\) would be open on \(T\), by the proposition that any path-connected topological component is open and closed on any locally path-connected topological space.
\(C_{c'}\) would be open on \(C'\), by the proposition that for any topological space and any topological subspace that is not necessarily open on the basespace, any subset of the subspace is open on the subspace if it is open on the basespace.
So, there would be the set of some open subsets of \(C'\), \(S := \{C_{c'} \vert c' \in C'\}\), in which \(C\) would be contained.
As the path-connected components were some equivalence classes, any 2 path-connected components would be the same or disjoint.
So, \(S \setminus \{C\}\) would be nonempty and \(C \cap \cup (S \setminus \{C\}) = \emptyset\).
\(\cup (S \setminus \{C\})\) would be open on \(C'\), because it would be the union of some open subsets.
\(C' = \cup S = C \cup \cup (S \setminus \{C\})\).
That would mean that \(C'\) was not connected, a contradiction: refer to the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger.
So, \(C = C'\).
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 for topological space, path-connected component is contained in connected component
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 path-connected component is contained in the corresponding connected component.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(C\): \(\in \{\text{ the path-connected components of } T\}\)
\(C'\): \(\in \{\text{ the connected components of } T\}\), such that \(c \in C'\) for any fixed \(c \in C\)
//
Statements:
\(C \subseteq C'\)
//
2: Note
\(C'\) is uniquely determined, by this proposition, because for any other \(\widetilde{C'}\) taken based on \(\widetilde{c} \in C\), \(C \subseteq \widetilde{C'}\) means that \(c \in \widetilde{C'}\), so, \(\widetilde{C'}\) is the connected component such that \(c \in \widetilde{C'}\), which is \(C'\).
3: Proof
Whole Strategy: Step 1: see that \(C\) is a path-connected subspace; Step 2: see that \(C\) is a connected subspace; Step 3: see that \(C \subseteq C'\).
Step 1:
\(C\) is a path-connected topological subspace, by the proposition that any path-connected topological component is exactly any path-connected topological subspace that cannot be made larger.
Step 2:
\(C\) is a connected topological subspace, by the proposition that any path-connected topological space is connected.
Step 3:
So, for each \(c' \in C\), \(c\) and \(c'\) are connected, because they are contained in the connected subspace, \(C\), which means that \(C \subseteq C'\).
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 path-connected topological space is connected
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 path-connected topological space is connected.
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:
\(T\): \(\in \{\text{ the path-connected topological spaces }\}\)
//
Statements:
\(T \in \{\text{ the connected topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(T\) was not connected, and find a contradiction.
Step 1:
Let us suppose that \(T\) was not connected.
There would be some nonempty open subsets, \(U_1, U_2 \subseteq T\), such that \(T = U_1 \cup U_2\) and \(U_1 \cap U_2 = \emptyset\).
Let \(t_1 \in U_1\) and \(t_2 \in U_2\) be any, which would be possible, because \(U_1\) and \(U_2\) were nonempty.
There would be a path, \(\gamma: [0, 1] \to T\), such that \(\gamma (0) = t_1\) and \(\gamma (1) = t_2\).
\(\gamma^{-1} (T) = [0, 1]\), by the proposition that the preimage of the whole codomain of any map is the whole domain.
But \(\gamma^{-1} (T) = \gamma^{-1} (U_1 \cup U_2) = \gamma^{-1} (U_1) \cup \gamma^{-1} (U_2)\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets, where \(\gamma^{-1} (U_1) \cap \gamma^{-1} (U_2) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint, \(\gamma^{-1} (U_1), \gamma^{-1} (U_2) \subseteq [0, 1]\) were open, because \(\gamma\) was continuous, and \(\gamma^{-1} (U_1)\) and \(\gamma^{-1} (U_2)\) were nonempty, because \(0 \in \gamma^{-1} (U_1)\) and \(1 \in \gamma^{-1} (U_2)\), which would mean that \([0, 1]\) was not connected.
But \([0, 1]\) was a connected topological space, by the proposition that the set of all the connected topological subspaces of the \(\mathbb{R}\) Euclidean topological space is the set of all the intervals, a contradiction.
So, \(T\) is connected.
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 connected component is quasi-connected component on locally connected topological space
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 connected component is the corresponding quasi-connected component on any locally connected topological space.
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:
\(T\): \(\in \{\text{ the locally connected topological spaces }\}\)
\(C\): \(\in \{\text{ the connected components of } T\}\)
\(C'\): \(\in \{\text{ the quasi-connected components of } T\}\), such that \(c \in C'\) for any fixed \(c \in C\)
//
Statements:
\(C = C'\)
//
2: Proof
Whole Strategy: Step 1: see that \(C \subseteq C'\) and \(C'\) is uniquely determined; Step 2: see that \(C'\) is open; Step 3: see that \(C'\) as the topological subspace is quasi-connected and \(C'\) is connected; Step 4: conclude the proposition.
Step 1:
\(C \subseteq C'\), by the proposition that for any topological space, each connected component is contained in the corresponding quasi-connected component.
\(C'\) is uniquely determined, because \(C'\) is an equivalence class and is a part of the division of \(T\).
Step 2:
\(C' \subseteq T\) is open, by the proposition that any quasi-connected component is open on any locally connected topological space.
Step 3:
Let us see that \(C'\) as the topological subspace is quasi-connected.
Note that \(C'\)'s being a quasi-connected component is different from \(C'\)'s being a quasi-connected topological space in general, because for the former, we think of the set of the continuous maps from \(T\) into any discrete topological spaces while for the latter, we think of the set of the continuous maps from \(C'\) into any discrete topological spaces.
Step 3 Strategy: Step 3-1: take the set of the continuous maps from \(C'\) into any discrete topological spaces, \(F\), and for each \(f: C' \to T' \in F\), extend \(f\) to a continuous \(f': T \to T'\); Step 3-2: see that for each \(c'_1, c'_2 \in C'\), \(f (c'_1) = f' (c'_1) = f' (c'_2) = f (c'_2)\).
Step 3-1:
Let us take the set of the continuous maps from \(C'\) into any discrete topological spaces, \(F\).
Let \(f: C' \to T' \in F\) be any.
Let \(t' \in T'\) be any.
Let us take the extension of \(f\), \(f': T \to T', t \mapsto f (t) \text{ when } t \in C'; \mapsto t' \text{ otherwise }\).
\(f'\) is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous: \(\{C', T \setminus C'\}\) is an open cover of \(T\), by the proposition that for any topological space, each quasi-connected component is closed, and \(f' \vert_{C'}: C' \to T' = f\) is continuous and \(f' \vert_{T \setminus C'}: T \setminus C' \to T'\) is continuous, because it is constant, by the proposition that any constant map between any topological spaces is continuous.
Step 3-2:
For each \(c'_1, c'_2 \in C'\), \(f (c'_1) = f' (c'_1) = f' (c'_2) = f (c'_2)\), because \(C'\) is a quasi-connected component.
That means that \(C'\) is a quasi-connected topological subspace.
Step 4:
So, \(C'\) is a connected topological subspace, by the proposition that any quasi-connected topological space is connected.
So, \(C' \subseteq C\), because each pair of points on \(C'\) is connected, so, all the points of \(C'\) are in the same connected component.
So, \(C = C'\).
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 constant map between topological spaces is continuous
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 constant map between any topological spaces is continuous.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\)
//
Statements:
\(\exists t_0 \in T_2 (\forall t \in T_1 (f (t) = t_0))\)
\(\implies\)
\(f \in \{\text{ the continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: see that for each \(t \in T_1\), \(f\) is continuous at \(t\).
Step 1:
Let \(t \in T_1\) be any.
\(f (t) = t_0\).
Let \(U_{t_0} \subseteq T_2\) be any open neighborhood of \(t_0\) on \(T_2\).
\(T_1\) is an open neighborhood of \(t\) on \(T_1\) and \(f (T_1) = \{t_0\} \subseteq U_{t_0}\).
So, \(f\) is continuous at \(t\).
As \(t\) is arbitrary, \(f\) is continuous.
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 quasi-connected topological space is connected
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 quasi-connected topological space is connected.
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:
\(T\): \(\in \{\text{ the quasi-connected topological spaces }\}\)
//
Statements:
\(T \in \{\text{ the connected topological spaces }\}\)
//
2: Note
In general, a topological quasi-connected component of \(T\) is not any topological connected component of \(T\) (although it contains some topological connected components), but when the topological quasi-connected component is \(T\), it is the connected component, by this proposition.
3: Proof
Whole Strategy: Step 1: suppose that \(T\) was not connected, and find a contradiction; Step 2: as an aside, see that the same logic does not work for when a topological quasi-connected component is not \(T\).
Step 1:
Let us suppose that \(T\) was not connected.
\(T = U_1 \cup U_2\) where \(U_1, U_2 \subseteq T\) were some nonempty open subsets of \(T\) such that \(U_1 \cap U_2 = \emptyset\).
Let us take the discrete \(T' = \{1, 2\}\) and \(f \in F: T \to T', t \mapsto 1 \text{ when } t \in U_1; \mapsto 2 \text{ when } t \in U_2\), which would be indeed continuous, because \(f^{-1} (\emptyset) = \emptyset\), \(f^{-1} (\{1\}) = U_1\), \(f^{-1} (\{2\}) = U_2\), and \(f^{-1} (T') = T\).
Then, there would be a \(u_1 \in U_1\) and a \(u_2 \in U_2\) such that \(f (u_1) = 1 \neq 2 = f (u_2)\), a contradiction against that \(T\) was a topological quasi-connected component: \(u_1\) and \(u_2\) were not topological quasi-connected.
So, \(T\) is connected.
Step 2:
As an aside, let us see why the same logic does not work for when a topological quasi-connected component is not \(T\).
Let \(C \subset T\) be a topological quasi-connected component.
Supposing \(C = U_1 \cup U_2\), taking the discrete \(T' = \{1, 2\}\) and \(f: C \to T', c \mapsto 1 \text{ when } c \in U_1; \mapsto 2 \text{ when } c \in U_2\), \(f\) would not be any map from \(T\) but from \(C\), while \(f\) needed to be a continuous map from \(T\), so, \(f\) would need to be able to be extended on the domain, \(T\), to be continuous, which cannot be proved.
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>
definition of quasi-connected topological space
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of quasi-connected topological space.
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:
\(*T\): \(\in \{\text{ the topological spaces }\}\)
\( R\): \(= \text{ the quasi-connectedness of } 2 \text{ points on } T\)
//
Conditions:
\(\forall t_1, t_2 \in T (t_1 R t_2)\)
//
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 quasi-connected component is open on locally connected topological space
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 quasi-connected component is open on any locally connected topological space.
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:
\(T\): \(\in \{\text{ the locally connected topological spaces }\}\)
\(C'\): \(\in \{\text{ the quasi-connected components of } T\}\)
//
Statements:
\(C' \in \{\text{ the open subsets of } T\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(C' = \cup_{c' \in C'} C_{c'}\) where \(C_{c'}\) is the connected component that contains \(c'\); Step 2: see that \(C_{c'}\) is open; Step 3: conclude the proposition.
Step 1:
\(C' = \cup_{c' \in C'} C_{c'}\) where \(C_{c'}\) is the connected component that contains \(c'\), because \(C_{c'} \subseteq C'\), by the proposition that for any topological space, each connected component is contained in the corresponding quasi-connected component.
Step 2:
Each \(C_{c'} \subseteq T\) is open, by the proposition that for any locally connected topological space, each connected component is open.
Step 3:
So, \(C' = \cup_{c' \in C'} C_{c'}\) is open on \(T\) as the union of some open subsets.
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 topological quasi-connected component is closed
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, each quasi-connected component is closed.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(C\): \(\in \{\text{ the quasi-connected components of } T\}\)
//
Statements:
\(C \in \{\text{ the closed subsets of } T\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(c \in C\), and see that \(C = \cap_{f \in F} f^{-1} (f (c))\).
Step 1:
Let \(c \in C\) be any.
\(C = \{t \in T \vert \forall f \in F (f (t) = f (c))\}\).
\(C = \cap_{f \in F} f^{-1} (\{f (c))\}\), because for each \(t \in \{t \in T \vert \forall f \in F (f (t) = f (c))\}\), for each \(f \in F\), \(f (t) = f (c)\), so, \(t \in f^{-1} (f (c))\) for each \(f \in F\), so, \(t \in \cap_{f \in F} f^{-1} (\{f (c))\})\); for each \(t \in \cap_{f \in F} f^{-1} (\{f (c))\})\), \(t \in f^{-1} (\{f (c))\})\) for each \(f \in F\), so, \(f (t) = f (c)\) for each \(f \in F\), so, \(t \in \{t \in T \vert \forall f \in F (f (t) = f (c))\}\).
\(\{f (c)\} \subseteq T'\) is closed, because \(T' \setminus \{f (c)\} \subseteq T'\) is open.
So, \(f^{-1} (\{f (c)) \subseteq T\) is closed, by the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
\(\cap_{f \in F} f^{-1} (\{f (c))\} \subseteq T\) is closed, by the proposition that the intersection of any possibly uncountable number of closed sets or the union of any finite number of closed sets is closed.
So, \(C \subseteq T\) is closed.
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, connected component is contained in quasi-connected component
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, each connected component is contained in the corresponding quasi-connected component.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(C\): \(\in \{\text{ the connected components of } T\}\)
\(C'\): \(\in \{\text{ the quasi-connected components of } T\}\), such that \(c \in C'\) for any fixed \(c \in C\)
//
Statements:
\(C \subseteq C'\)
//
For any fixed \(c \in C\), there is the determined \(C'\) such that \(c \in C'\), and such \(C'\) does not depend on \(c\), by this proposition.
2: Proof
Whole Strategy: Step 1: take any fixed \(c \in C\) and the determined \(C'\) such that \(c \in C'\); Step 2: see that for each \(c' \in C\), for each \(f \in F\), \(f (c') = f (c)\).
Step 1:
Let us take any fixed \(c \in C\).
There is the determined \(C'\) such that \(c \in C'\), which is indeed determined, because \(C'\) is an equivalence class and any set of equivalence classes is a division of the set.
Step 2:
Let \(f \in F: T \to T'\) be any.
\(T = f^{-1} (T')\), by the proposition that the preimage of the whole codomain of any map is the whole domain, \(= f^{-1} (\{f (c)\} \cup (T' \setminus \{f (c)\})) = f^{-1} (\{f (c)\}) \cup f^{-1} (T' \setminus \{f (c)\})\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
\(f^{-1} (\{f (c)\}) \cap f^{-1} (T' \setminus \{f (c)\}) = \emptyset\), by the proposition that the preimages of any disjoint subsets under any map are disjoint.
\(f^{-1} (\{f (c)\})\) and \(f^{-1} (T' \setminus \{f (c)\})\) are some open subsets of \(T\), because \(T'\) is discrete and \(f\) is continuous.
\(C = T \cap C = (f^{-1} (\{f (c)\}) \cup f^{-1} (T' \setminus \{f (c)\})) \cap C = (f^{-1} (\{f (c)\}) \cap C) \cup (f^{-1} (T' \setminus \{f (c)\}) \cap C)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
\(f^{-1} (\{f (c)\}) \cap C\) and \(f^{-1} (T' \setminus \{f (c)\}) \cap C\) are some open subsets of \(C\) as the topological subspace and \((f^{-1} (\{f (c)\}) \cap C) \cap (f^{-1} (T' \setminus \{f (c)\}) \cap C) = \emptyset\), because \(f^{-1} (\{f (c)\}) \cap f^{-1} (T' \setminus \{f (c)\}) = \emptyset\).
So, as \(C\) is a connected topological subspace of \(T\), by the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger, \(f^{-1} (T' \setminus \{f (c)\}) \cap C = \emptyset\): \(f^{-1} (\{f (c)\}) \cap C \neq \emptyset\), because \(c \in f^{-1} (\{f (c)\}) \cap C\).
That means that \(f (C) \subseteq \{f (c)\}\), which means that for each \(c' \in C\), \(f (c') = f (c)\).
That means that \(C \subseteq C'\).
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>
definition of topological quasi-connected component
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of topological quasi-connected component.
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:
\( T\): \(\in \{\text{ the topological spaces }\}\)
\( R\): \(= \text{ the quasi-connectedness of } 2 \text{ points on } T\)
\( T / R\): \(= \text{ the set of the equivalence classes of } T \text{ by } R\)
\(*C\): \(\in T / R\)
//
Conditions:
//
2: Note
This is called "quasi-connected component", because each connected component of \(T\) is contained in a quasi-connected component of \(T\), but not necessarily vice versa, by the proposition that for any topological space, each connected component is contained in the corresponding quasi-connected component.
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 topological quasi-connectedness of 2 points is equivalence relation on topological space
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 topological quasi-connectedness of 2 points is an equivalence relation on any topological space.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(R\): \(= \text{ the quasi-connectedness of } 2 \text{ points on } T\)
//
Statements:
\(R \in \{\text{ the equivalence relations }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(R\) satisfies the conditions to be an equivalence relation.
Step 1:
1) \(\forall t \in T (t \sim t)\): reflexivity: for each \(f \in F\), \(f (t) = f (t)\).
2) \(\forall t_1, t_2 \in T (t_1 \sim t_2 \implies t_2 \sim t_1)\): symmetry: for each \(f \in F\), \(f (t_1) = f (t_2)\), so, for each \(f \in F\), \(f (t_2) = f (t_1)\).
3) \(\forall t_1, t_2, t_3 \in T ((t_1 \sim t_2 \land t_2 \sim t_3)\implies t_1 \sim t_3)\): transitivity: for each \(f \in F\), \(f (t_1) = f (t_2)\) and \(f (t_2) = f (t_3)\), so, for each \(f \in F\), \(f (t_1) = f (t_3)\).
So, \(R\) is an equivalence relation.
References
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definition of topological quasi-connectedness of 2 points
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of topological quasi-connectedness of 2 points.
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:
\( T\): \(\in \{\text{ the topological spaces }\}\)
\( F\): \(= \{f: T \to T' \vert T' \in \{\text{ the discrete topological spaces }\}, f \in \{\text{ the continuous maps }\}\}\)
\(*R\): \(\in \{\text{ the relations on } T\}\)
//
Conditions:
\(\forall t_1, t_2 \in T (t_1 R t_2 \iff \forall f \in F (f (t_1) = f (t_2)))\)
//
2: Note
This is called "quasi-connected", because if \(t_1\) and \(t_2\) are connected, they are quasi-connected, but not necessarily vice versa, by the proposition that for any topological space, each connected component is contained in the corresponding quasi-connected component.
References
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definition of discrete topological space
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of discrete topological space.
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:
\(*T\): \(\in \{\text{ the topological spaces }\}\) with any topology, \(O\)
//
Conditions:
\(\forall t \in T (\{t\} \in O)\)
//
References
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definition of set of equivalence classes of set by equivalence relation
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of set of equivalence classes of set by equivalence relation.
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:
\( S\): \(\in \{\text{ the sets }\}\)
\( R\): \(\in \{\text{ the equivalence relations on } S\}\)
\(*S / R\): \(\subseteq Pow (S)\), \(= \{S_s = \{s' \in S \vert s R s'\} \subseteq S \vert s \in S\}\)
//
Conditions:
//
2: Note
That does not mean that \(S / R\) has \(\vert S \vert\) elements, because \(S_{s_1}\) may equal \(S_{s_2}\).
Let us see that \(S / R\) is a division of \(S\), which means that \(S / R\) covers \(S\) and the elements of \(S / R\) are disjoint.
For each \(s \in S\), \(s \in S_s\), so, \(S / R\) covers \(S\).
Let us see that the elements of \(S / R\) are disjoint.
Let \(S_{s_1}, S_{s_2} \in S / R\) be any such that \(S_{s_1} \neq S_{s_2}\).
If \(S_{s_1} \cap S_{s_2} \neq \emptyset\), there would be an \(s \in S_{s_1} \cap S_{s_2}\), and for each \(s' \in S_{s_2}\), \(s_1 R s\), because \(s \in S_{s_1}\), \(s R s_2\), because \(s \in S_{s_2}\), \(s_2 R s'\), because \(s' \in S_{s_2}\), and so, \(s_1 R s'\), which would mean that \(s' \in S_{s_1}\), so, \(S_{s_2} \subseteq S_{s_1}\), and \(S_{s_1} \subseteq S_{s_2}\), by the symmetry, so, \(S_{s_1} = S_{s_2}\), a contradiction, so, \(S_{s_1} \cap S_{s_2} = \emptyset\).
So, the elements of \(S / R\) are disjoint.
So, \(S / R\) is a division of \(S\).
References
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