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description/proof of that for topological space with equivalence relation and subspace with subset equivalence relation, canonical injection from quotient topological space of subspace into quotient topological space of space 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 for any topological space with any equivalence relation and any subspace with the subset equivalence relation, the canonical injection from the quotient topological space of the subspace into the quotient topological space of the space 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:
\(T'\): \(\in \{\text{ the topological spaces }\}\), with any equivalence relation, \(\sim'\)
\(T\): \(\subseteq T'\), with the subset equivalence relation, \(\sim\)
\(f'\): \(: T' \to T' / \sim', t' \mapsto [t']'\)
\(f\): \(: T \to T / \sim, t \mapsto [t]\)
\(T' / \sim'\): \(= \text{ the quotient set with the quotient topology with respect to } f'\)
\(T / \sim\): \(= \text{ the quotient set with the quotient topology with respect to } f\)
\(g\): \(: T / \sim \to T' / \sim', [t] \mapsto [t]'\)
//
Statements:
\(g \in \{\text{ the continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f' \vert_T = g \circ f\); Step 2: see that for each open \(U' \subseteq T' / \sim'\), \({f' \vert_T}^{-1} (U') = (g \circ f)^{-1} (U') = f^{-1} (g^{-1} (U'))\), and see that \(g^{-1} (U')\) is open.
Step 1:
\(g\) is valid and \(f' \vert_T = g \circ f\), by the proposition that for any set with any equivalence relation and any subset with the subset equivalence relation, there is the canonical injection from the quotient set of the subset into the quotient set of the set.
Step 2:
Let \(U' \subseteq T' / \sim'\) be any open subset.
\({f' \vert_T}^{-1} (U') = (g \circ f)^{-1} (U')\).
\({f' \vert_T}^{-1} (U') = f'^{-1} (U') \cap T\), by the proposition that for any map between sets and its any domain-restriction, the preimage under the domain-restricted map is the intersection of the preimage under the original map and the restricted domain, which is open on \(T\), because \(f'^{-1} (U') \subseteq T'\) is open, because \(f'\) is continuous with \(T' / \sim'\) given the quotient topology.
\((g \circ f)^{-1} (U') = f^{-1} (g^{-1} (U'))\), by the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order.
So, \(f^{-1} (g^{-1} (U')) \subseteq T\) is open.
Then, \(g^{-1} (U') \subseteq T / \sim\) is open, by the definition of quotient topology.
So, \(g\) is continuous.
References
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description/proof of that for set with equivalence relation and subset with subset equivalence relation, there is canonical injection from quotient set of subset into quotient set of set
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 set with any equivalence relation and any subset with the subset equivalence relation, there is the canonical injection from the quotient set of the subset into the quotient set of the set.
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:
\(S'\): \(\in \{\text{ the sets }\}\), with any equivalence relation, \(\sim'\)
\(S\): \(\subseteq S'\), with the subset equivalence relation, \(\sim\)
\(f'\): \(: S' \to S' / \sim', s' \mapsto [s']'\)
\(f\): \(: S \to S / \sim, s \mapsto [s]\)
\(g\): \(: S / \sim \to S' / \sim', [s] \mapsto [s]'\)
//
Statements:
\(g \in \{\text{ the injections }\}\)
\(\land\)
\(f' \vert_S = g \circ f\)
//
2: Proof
Whole Strategy: Step 1: see that \(g\) is indeed well-defined; Step 2: see that for each \([s_1], [s_2] \in S / \sim\) such that \([s_1] \neq [s_2]\), \(g ([s_1]) \neq g ([s_2])\); Step 3: see that \(f' \vert_S = g \circ f\).
Step 1:
Let us see that \(g\) is indeed well-defined.
For each \([s] \in S / \sim\), let \(s_1 \in S\) be any other that \([s_1] = [s]\), then, \(s_1 \sim s\), which implies that \(s_1 \sim' s\), so, \([s_1]' = [s]'\), so, \(g ([s]) = [s]'\) does not depend on the choice of the representative of \([s]\).
So, \(g\) is well-defined.
Step 2:
Let \([s_1], [s_2] \in S / \sim\) be any such that \([s_1] \neq [s_2]\).
\(g ([s_1]) \neq g ([s_2])\), because if \(g ([s_1]) = [s_1]' = [s_2]' = g ([s_2])\), which would imply that \(s_1 \sim' s_2\), so, \(s_1 \sim s_2\), so, \([s_1] = [s_2]\), a contradiction.
So, \(g\) is an injection.
Step 3:
For each \(s \in S\), \(f' \vert_S (s) = [s]'\), while \(g \circ f (s) = g ([s]) = [s]'\).
So, \(f' \vert_S = g \circ f\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
definition of subset equivalence relation
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of subset 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:
\( S'\): \(\in \{\text{ the sets }\}\), with any equivalence relation, \(\sim'\)
\( S\): \(\subseteq S'\)
\(*\sim\): \(\in \{\text{ the equivalence relations on } S\}\)
//
Conditions:
\(\forall s_1, s_2 \in S (s_1 \sim s_2 \iff s_1 \sim' s_2)\)
//
2: Note
Let us see that \(\sim\) is indeed an equivalence relation on \(S\).
\(\sim\) is indeed a relation, because \(\sim = \{\langle s_1, s_2 \rangle \in S \times S \vert s_1 \sim s_2\}\), which is well-defined, because \(s_1 \sim s_2\) if and only if \(s_1 \sim' s_2\), which equals \(\langle s_1, s_2 \rangle \in \sim'\), so, \(\sim = \{\langle s_1, s_2 \rangle \in S \times S \vert \langle s_1, s_2 \rangle \in \sim'\}\).
1) \(\forall s \in S (s \sim s)\): reflexivity: \(s \sim' s\).
2) \(\forall s_1, s_2 \in S (s_1 \sim s_2 \implies s_2 \sim s_1)\): symmetry: \(s_1 \sim s_2 \implies s_1 \sim' s_2 \implies s_2 \sim' s_1 \implies s_2 \sim s_1\).
3) \(\forall s_1, s_2, s_3 \in S ((s_1 \sim s_2 \land s_2 \sim s_3) \implies s_1 \sim s_3)\): transitivity: \((s_1 \sim s_2 \land s_2 \sim s_3) \implies (s_1 \sim' s_2 \land s_2 \sim' s_3) \implies s_1 \sim' s_3 \implies s_1 \sim s_3\).
So, \(\sim\) is an equivalence relation on \(S\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that finite composition of quotient maps is quotient, if codomains of constituent maps equal domains of succeeding maps
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 finite composition of quotient maps is quotient, if the codomains of the constituent maps equal the domains of the succeeding maps.
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:
\(\{T_1, ..., T_{n + 1}\}\): \(\subseteq \{\text{ the topological spaces }\}\)
\(\{f_1: T_1 \to T_2, ..., f_n: T_n \to T_{n + 1}\}\): \(\subseteq \{\text{ the quotient maps }\}\)
\(f_n \circ ... \circ f_1\): \(: T_1 \to T_{n + 1}\)
//
Statements:
\(f_n \circ ... \circ f_1 \in \{\text{ the quotient maps }\}\)
//
2: Note
The requirement that "the codomains of the constituent maps equal the domains of the succeeding maps" is imperative, because otherwise, \(f_n \circ ... \circ f_1\) is not even guaranteed to be surjective, by the proposition that a finite composition of surjections is not necessarily any surjection, and no non-surjective map can be quotient.
3: Proof
Whole Strategy: Step 1: see that \(f_n \circ ... \circ f_1\) is a continuous surjection; Step 2: see that for each \(S \subseteq T_{n + 1}\) such that \((f_n \circ ... \circ f_1)^{-1} (S) \subseteq T_1\) is open, \(S\) is open.
Step 1:
\(f_n \circ ... \circ f_1\) is continuous, by the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.
\(f_n \circ ... \circ f_1\) is a surjection, by the proposition that any finite composition of surjections is a surjection, if the codomains of the constituent surjections equal the domains of the succeeding surjections.
Step 2:
Let \(S \subseteq T_{n + 1}\) be any such that \((f_n \circ ... \circ f_1)^{-1} (S) \subseteq T_1\) is open.
\((f_n \circ ... \circ f_1)^{-1} (S) = f_1^{-1} (f_2^{-1} (... f_{n - 1}^{-1} (f_n^{-1} (S) \cap S_n) ...) \cap S_2))\), by the proposition that for any maps composition, the preimage under the composition is the composition of the map preimages in the reverse order, \(= f_1^{-1} (f_2^{-1} (... f_{n - 1}^{-1} (f_n^{-1} (S)) ...)))\): "\(\cap S_n\)", e.t.c. are unnecessary, because \(S'_j = S_j\) in this case.
As it is open, \(f_2^{-1} (... f_{n - 1}^{-1} (f_n^{-1} (S)) ...) \subseteq T_2\) is open, because \(f_1\) is quotient, ..., \(f_n^{-1} (S) \subseteq T_n\) is open, because \(f_{n - 1}\) is quotient, and \(S\) is open, because \(f_n\) is quotient.
So, \(f_n \circ ... \circ f_1\) is quotient.
References
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description/proof of that for \(1\)-dimensional Euclidean topological space and equivalence relation that \(2\) elements are equivalent iff their difference is rational, quotient topology is trivial
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 the \(1\)-dimensional Euclidean topological space and the equivalence relation that any \(2\) elements are equivalent if and only if their difference is rational, the quotient topology is trivial.
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:
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(\sim\): \(\in \{\text{ the equivalence relations on } \mathbb{R}\}\), such that for each \(r_1, r_2 \in \mathbb{R}\), \(r_1 \sim r_2 \iff r_1 - r_2 \in \mathbb{Q}\)
\(\mathbb{R} / \sim\): \(= \text{ the quotient set }\) with the quotient topology with respect to \(f\)
\(f\): \(: \mathbb{R} \to \mathbb{R} / \sim\), such that for each \(r \in \mathbb{R}\), \(r \in f (r)\)
\(\)
//
Statements:
\(\mathbb{R} / \sim \in \{\text{ the trivial topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: for any nonempty open \(U \subseteq \mathbb{R} / \sim\), see that for each \(r' \in \mathbb{R}\), \(r' \in f^{-1} (U)\), by taking any \(r \in f^{-1} (U)\) and \(B_{r, \epsilon} \subseteq f^{-1} (U)\) and seeing that \(r' - q \in B_{r, \epsilon}\) for a \(q \in \mathbb{Q}\).
Step 1:
Let \(U \subseteq \mathbb{R} / \sim\) be any nonempty open subset.
Let \(r \in f^{-1} (U)\) be any, which exists, because \(U\) is nonempty and \(f\) is a surjection.
There is a \(B_{r, \epsilon} \subseteq \mathbb{R}\) such that \(B_{r, \epsilon} \subseteq f^{-1} (U)\), because \(f^{-1} (U) \subseteq \mathbb{R}\) is open.
Let \(r' \in \mathbb{R}\) be any.
\(r' - r - \epsilon \lt r' - r + \epsilon\).
So, there is a \(q \in \mathbb{Q}\) such that \(r' - r - \epsilon \lt q \lt r' - r + \epsilon\), by Note for the way for systematically choosing a rational number that is larger than any real number and is equal to or smaller than another any real number.
\(- r' + r - \epsilon \lt - q \lt - r' + r + \epsilon\), \(r - \epsilon \lt r' - q \lt r + \epsilon\), so, \(r' - q \in B_{r, \epsilon}\), so, \(r' - q \in f^{-1} (U)\).
\(f (r') = f (r' - q) \in U\), so, \(r' \in f^{-1} (U)\).
So, \(f^{-1} (U) = \mathbb{R}\).
So, \(U = \mathbb{R} / \sim\), because if \([r] \notin U\), \(r \notin f^{-1} (U)\), a contradiction.
That means that the open subsets of \(\mathbb{R} / \sim\) are \(\emptyset\) and \(\mathbb{R} / \sim\).
So, \(\mathbb{R} / \sim\) is trivial.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for real numbers set and equivalence relation that \(2\) elements are equivalent iff their difference is rational, quotient set is uncountable
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 the real numbers set and the equivalence relation that any \(2\) elements are equivalent if and only if their difference is rational, the quotient set is uncountable.
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:
\(\mathbb{R}\):
\(\sim\): \(\in \{\text{ the equivalence relations on } \mathbb{R}\}\), such that for each \(r_1, r_2 \in \mathbb{R}\), \(r_1 \sim r_2 \iff r_1 - r_2 \in \mathbb{Q}\)
\(\mathbb{R} / \sim\): \(= \text{ the quotient set }\)
//
Statements:
\(\mathbb{R} / \sim \notin \{\text{ the countable sets }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(\sim\) is indeed an equivalence relation; Step 2: take the classification map, \(g: \mathbb{R} \to \mathbb{R} / \sim\), any representatives map, \(f: \mathbb{R} / \sim \to \mathbb{R}\), and the map, \(h: \mathbb{R} \to (\mathbb{R} / \sim) \times \mathbb{Q}, r \mapsto (g (r), r - f (g (r)))\), and see that \(h\) is an injection.
Step 1:
Let us see that \(\sim\) is indeed an equivalence relation.
1) \(\forall r \in \mathbb{R} (r \sim r)\): reflexivity: \(r - r = 0 \in \mathbb{Q}\), so, \(r \sim r\).
2) \(\forall r_1, r_2 \in \mathbb{R} (r_1 \sim r_2 \implies r_2 \sim r_1)\): symmetry: \(r_1 - r_2 \in \mathbb{Q}\), so, \(r_2 - r_1 = - (r_1 - r_2) \in \mathbb{Q}\), so, \(r_2 \sim r_1\).
3) \(\forall r_1, r_2, r_3 \in \mathbb{R} ((r_1 \sim r_2 \land r_2 \sim r_3)\implies r_1 \sim r_3)\): transitivity: \(r_1 - r_2 \in \mathbb{Q}\) and \(r_2 - r_3 \in \mathbb{Q}\), so, \(r_1 - r_3 = r_1 - r_2 + r_2 - r_3 = (r_1 - r_2) + (r_2 - r_3) \in \mathbb{Q}\), so, \(r_1 \sim r_3\).
So, \(\sim\) is an equivalence relation.
Step 2:
There is the classification map, \(g: \mathbb{R} \to \mathbb{R} / \sim, r \mapsto [r]\), which is a surjection, as is mentioned in Note for the definition of quotient set.
There is a representatives map, \(f: \mathbb{R} / \sim \to \mathbb{R}\), which is an injection, as is mentioned in Note for the definition of representatives set of quotient set.
Let us define the map, \(h: \mathbb{R} \to (\mathbb{R} / \sim) \times \mathbb{Q}, r \mapsto (g (r), r - f (g (r)))\).
That is indeed valid, because as \(r, f (g (r)) \in g (r)\), \(r \sim f (g (r))\), so, \(r - f (g (r)) \in \mathbb{Q}\).
\(h\) is injective, because for each \(r_1, r_2 \in \mathbb{R}\) such that \(r_1 \neq r_2\), when \(g (r_1) \neq g (r_2)\), \(h (r_1) \neq h (r_2)\), and otherwise, \(f (g (r_1)) = f (g (r_2))\), so, \(r_1 - f (g (r_1)) \neq r_2 - f (g (r_2))\), because otherwise, \(r_1 = r_2\), a contradiction.
If \(\mathbb{R} / \sim\) was countable, \((\mathbb{R} / \sim) \times \mathbb{Q}\) would be countable, by the proposition that any finite product of countable sets is countable, so, there would be a bijection, \(h': (\mathbb{R} / \sim) \times \mathbb{Q} \to \mathbb{N}\), and \(h' \circ h: \mathbb{R} \to (\mathbb{R} / \sim) \times \mathbb{Q} \to \mathbb{N}\) would be an injection, by the proposition that any finite composition of injections is an injection, then, \(\mathbb{R}\) would be countable, by the proposition that for any infinite set, if there is any injection from the set into the natural numbers set, there is a bijection from the natural numbers set onto the set, a contradiction against that \(\mathbb{R}\) was not countable.
So, \(\mathbb{R} / \sim\) is not countable.
References
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description/proof of that infinite product of sets each of which has more than \(1\) elements is uncountable
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 any infinite product of sets each of which has more than \(1\) elements is uncountable.
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:
\(J\): \(\in \{\text{ the infinite index sets }\}\)
\(\{S_j \in \{\text{ the sets } \} \vert j \in J, 1 \lt \vert S_j \vert\}\):
//
Statements:
\(\times_{j \in J} S_j \notin \{\text{ the countable sets }\}\)
//
2: Note
When each \(S_j\) has only \(1\) element, \(\times_{j \in J} S_j\) is countable, in fact, has only \(1\) element, because for each \(f \in \times_{j \in J} S_j\), \(f (j)\) is the only element of \(S_j\) for each \(j \in J\), so, \(f\) is uniquely determined.
3: Proof
Whole Strategy: Step 1: suppose that there was a surjection, \(g: \mathbb{N} \to \times_{j \in J} S_j\), and find a contradiction.
Step 0:
Note that the axiom of choice is used without being mentioned explicitly where.
Step 1:
As \(J\) is infinite, there is an injection, \(g': \mathbb{N} \to J\): define it inductively as this: for \(0\), choose an element of \(J\); for \(1\), choose an element of the rest of \(J\), ..., and so on.
Let us suppose that there was a surjection, \(g: \mathbb{N} \to \times_{j \in J} S_j\).
Let \(f \in \times_{j \in J} S_j\) be as this: for each \(n \in \mathbb{N}\), \(f (g' (n)) \neq g (n) (g' (n))\), which would be possible, because \(S_{g' (n)}\) had more than \(1\) elements, so, while \(g (n) (g' (n))\) was determined, choose \(f (g' (n))\) from \(S_{g' (n)} \setminus \{g (n) (g' (n))\}\): as \(g'\) is injective, there is no duplication in \(\{g' (n) \vert n \in \mathbb{N}\}\); for each \(j \in J \setminus g' (\mathbb{N})\), take any element of \(S_{j}\) as \(f (j)\).
Then, \(f\) would not be covered by \(g\), because for each \(n \in \mathbb{N}\), \(f \neq g (n)\), because \(f (g' (n)) \neq g (n) (g' (n))\).
That is a contradiction against that \(g\) was surjective.
So, there is no surjection, \(g: \mathbb{N} \to \times_{j \in J} S_j\).
So, \(\times_{j \in J} S_j\) is not countable.
References
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description/proof of finite product of countable sets is countable
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 any finite product of countable sets is countable.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\), with \(\vert J \vert = n\) and any ordering
\(\{S_j \in \{\text{ the countable sets }\} \vert j \in J\}\):
//
Statements:
\(\times_{j \in J} S_j \in \{\text{ the countable sets }\}\)
//
2: Note
When \(J\) is infinite countable, \(\times_{j \in J} S_j\) is not countable in general, especially, when each \(S_j\) is infinite, \(\times_{j \in J} S_j\) is always not countable, by the proposition that any infinite product of sets each of which has more than \(1\) elements is uncountable.
3: Proof
Whole Strategy: Step 1: augment \(S_j\) to any infinite countable \(S'_j\) if necessary; Step 2: take a surjection, \(g: \mathbb{N} \setminus \{0\} \to \times_{j \in J} S'_j\).
Step 1:
For each \(j \in J\), if \(S_j\) is finite, let \(S_j\) be augmented to be infinite countable \(S'_j\), which is possible, because for example, \(S'_j := S_j \cup \mathbb{N}\) will do.
That we do because doing some special treatments for the case that some \(S_j\) s are finite is bothersome.
Step 2:
For each \(j \in J\), there is a bijection, \(g_j: \mathbb{N} \setminus \{0\} \to S'_j\).
Let us define a map, \(g: \mathbb{N} \setminus \{0\} \to \times_{j \in J} S'_j\), inductively as this.
For \(n\), let us take \(\{f' \in \times_{j \in J} S'_j \vert {g_{J_1}}^{- 1} \circ f' (J_1) + ... + {g_{J_n}}^{- 1} \circ f' (J_n) = n\}\), which has \(1\) element, because \({g_{J_1}}^{- 1} \circ f' (J_1) = ... = {g_{J_n}}^{- 1} \circ f' (J_n) = 1\) is the only possibility.
Then, let us define \(g (1)\) as the element.
Now, we have \(g \vert_{\{1, ..., m_n\}}\), where \(m_n = 1\).
For \(n + 1\), let us take \(\{f' \in \times_{j \in J} S'_j \vert {g_{J_1}}^{- 1} \circ f' (J_1) + ... + {g_{J_n}}^{- 1} \circ f' (J_n) = n + 1\}\), which has \(n\) elements, because \({g_{J_1}}^{- 1} \circ f' (J_1) = 2 \land {g_{J_2}}^{- 1} \circ f' (J_2) = ... = {g_{J_n}}^{- 1} \circ f' (J_n) = 1, ..., {g_{J_1}}^{- 1} \circ f' (J_1) = ... = {g_{J_{n - 1}}}^{- 1} \circ f' (J_{n - 1}) = 1 \land {g_{J_n}}^{- 1} \circ f' (J_n) = 2\) are the only possibilities.
Let us order the elements in the lexical order of \(({g_{J_1}}^{- 1} \circ f' (J_1), ..., {g_{J_n}}^{- 1} \circ f' (J_n))\), which means that for each \(f'_1, f'_2\), if \({g_{J_1}}^{- 1} \circ f'_1 (J_1) \neq {g_{J_1}}^{- 1} \circ f'_2 (J_1)\), the order of \(f'_1, f'_2\) is determined accordingly, otherwise, if \({g_{J_2}}^{- 1} \circ f'_1 (J_2) \neq {g_{J_2}}^{- 1} \circ f'_2 (J_2)\), the order of \(f'_1, f'_2\) is determined accordingly, and so on.
Then, let us define \(g (m_n + 1), ..., g (m_n + n)\) as the ordered elements.
Now, we have \(g \vert_{\{1, ..., m_{n + 1}\}}\), where \(m_{n + 1} = n + 1\).
Let us suppose that for \(n' - 1\), we have \(g \vert_{\{1, ..., m_{n' - 1}\}}\).
For \(n'\), let us take \(\{f' \in \times_{j \in J} S'_j \vert {g_{J_1}}^{- 1} \circ f' (J_1) + ... + {g_{J_n}}^{- 1} \circ f' (J_n) = n'\}\).
One does not bother to count the number of the elements, but it is certainly finite, because it is smaller than \(n'^n\), because each \({g_{J_j}}^{- 1} \circ f (J_j)\) can have only \(1, ..., n'\), and what is important is that the set of the elements is uniquely determined and finite, not to show the explicit formula of the number.
Let us order the set of the elements in the lexical order of \(({g_{J_1}}^{- 1} \circ f' (J_1), ..., {g_{J_n}}^{- 1} \circ f' (J_n))\).
Then, let us define \(g (m_{n' - 1} + 1), ..., g (m_{n'})\) as the ordered elements.
Now, we have \(g \vert_{\{1, ..., m_{n'}\}}\).
Thus, \(g\) has been defined.
\(g\) is a surjection, because for each \(f' \in \times_{j \in J} S'_j\), \({g_{J_1}}^{- 1} \circ f' (J_1) + ... + {g_{J_n}}^{- 1} \circ f' (J_n)\) has a definite value equal to or larger than \(n\), so, \(f\) is covered by the step for \({g_{J_1}}^{- 1} \circ f' (J_1) + ... + {g_{J_n}}^{- 1} \circ f' (J_n)\).
Let us define \(g': \times_{j \in J} S'_j \to \times_{j \in J} S_j\) such that for each \(f' \in \times_{j \in J} S'_j\), when \(f' (j) \in S_j\) for each \(j \in J\), \(g' (f')\) is the one such that \(g' (f') (j) = f' (j)\), and otherwise, \(g' (f')\) is the element of \(\times_{j \in J} S_j\) such that \(({g_{J_1}}^{- 1} \circ f (J_1), ..., {g_{J_n}}^{- 1} \circ f (J_n)) = (1, ..., 1)\).
\(g'\) is a surjection, because for each \(f \in \times_{j \in J} S_j\), there is the corresponding element of \(\times_{j \in J} S'_j\), which is mapped to \(f\).
\(g' \circ g: \mathbb{N} \setminus \{0\} \to \times_{j \in J} S_j\) is a surjection, by the proposition that any finite composition of surjections is a surjection, if the codomains of the constituent surjections equal the domains of the succeeding surjections
When \(\times_{j \in J} S_j\) is finite, \(\times_{j \in J} S_j\) is countable.
Otherwise, there is a bijection, \(g'': \mathbb{N} \setminus \{0\} \to \times_{j \in J} S_j\), by the proposition that for any infinite set, if there is a surjection from the natural numbers set onto the set, there is a bijection from the natural numbers set onto the set, so, \(\times_{j \in J} S_j\) is countable.
So, \(\times_{j \in J} S_j\) is countable, anyway.
References
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description/proof of that finite product set is 'sets - maps' isomorphic to sequential products of sets
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 any finite product set is 'sets - maps' isomorphic to the sequential products of the sets.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\), with \(\vert J \vert = n\) and any ordering
\(\{S_j \in \{\text{ the sets } \} \vert j \in J\}\):
\(\times_{j \in J} S_j\):
\(( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\):
//
Statements:
\(\times_{j \in J} S_j \cong_{sets} ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\)
//
2: Proof
Whole Strategy: Step 1: define a bijection, \(g: \times_{j \in J} S_j \to ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\).
Step 1:
Let us define the map, \(g: \times_{j \in J} S_j \to ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}, f \mapsto f_{n - 1}\), where \(f_1 \in S_{J_1} \times S_{J_2}\) (which is a map from \(\{1, 2\}\) into \(S_{J_1} \cup S_{J_2}\)) is determined as \(f_1 (1) = f (J_1)\) and \(f_1 (2) = f (J_2)\), \(f_2 \in (S_{J_1} \times S_{J_2}) \times S_{J_3}\) (which is a map from \(\{1, 2\}\) into \((S_{J_1} \times S_{J_2}) \cup S_{J_3}\)) is determined as \(f_2 (1) = f_1\) and \(f_2 (2) = f (J_3)\), ..., and \(f_{n - 1} \in ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\) (which is a map from \(\{1, 2\}\) into \(( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_{n - 1}} \cup S_{J_n}\)) is determined as \(f_{n - 1} (1) = f_{n - 1}\) and \(f_{n - 1} (2) = f (J_n)\).
\(g\) is an injection, because for any \(f, f' \in \times_{j \in J} S_j\) such that \(f \neq f'\), \(f (J_{n'}) \neq f' (J_{n'})\) for an \(n'\), and when \(n' = 1\), \(f_1 \neq f'_1\), because \(f_1 (1) = f (J_1) \neq f' (J_1) = f'_1 (1)\), and otherwise, \(f_{n' - 1} \neq f'_{n' - 1}\), because \(f_{n' - 1} (2) = f (J_{n'}) \neq f' (J_{n'}) = f'_{n' - 1} (2)\), then, \(f_{n - 1} \neq f'_{n - 1}\).
\(g\) is a surjection, because for each \(f_{n - 1} \in ( ... ((S_{J_1} \times S_{J_2}) \times S_{J_3}) \times ... ) \times S_{J_n}\), there is the \(f \in \times_{j \in J} S_j\) such that \(f (J_1) = f_1 (1)\), \(f (J_2) = f_1 (2)\), \(f (J_3) = f_2 (2)\), ..., \(f (J_n) = f_{n - 1} (2)\), and \(g (f) = f_{n - 1}\).
So, \(g\) is a bijection.
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of a way to systematically choose rational number that is larger than real number and is equal to or smaller than another real number
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of a way for systematically choosing a rational number that is larger than any real number and is equal to or smaller than another any real number.
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:
\(r_1\): \(\in \mathbb{R}\)
\(r_2\): \(\in \mathbb{R}\), such that \(r_1 \lt r_2\)
//
Statements:
choose \(q\) as follows:
let \(r_1\) and \(r_2\) be expressed as the decimals without any trailing '999...'
when \(0 \le r_2\), when \(r_1 \lt 0\), take \(q = 0\), otherwise, take the 1st digit on which \(r_1\) and \(r_2\) disagree and take \(q\) as \(r_2\) with with the subsequent digits cut off
when \(r_2 \lt 0\), take the 1st digit on which \(r_1\) and \(r_2\) disagree, take the 1st digit of \(r_2\) after that that is not '9', and take \(q\) as \(r_2\) with the digit incremented by \(1\) and the subsequent digits cut off
\(\implies\)
\(q\) has been systematically chosen satisfying \(q \in \mathbb{Q} \land r_1 \lt q \le r_2\)
//
2: Note
We cannot do like \(Max (\{q \in \mathbb{Q} \vert q \le r_2\})\), because such the maximum may not exist: what is the maximum when \(r_2 = \sqrt{2}\)? We cannot do like \(Sup (\{q \in \mathbb{Q} \vert q \le r_2\})\), because such the supremum may not exist in \(\mathbb{Q}\) but in \(\mathbb{R}\): what is the supremum in \(\mathbb{Q}\) when \(r_2 = \sqrt{2}\)?
When we have only some finite number of pairs, \((r_{1, 1}, r_{1, 2}), ..., (r_{n, 1}, r_{n, 2})\), we can just claim that we can choose some rational numbers, \(r_{1, 1} \lt q_1 \le r_{1, 2}, ..., r_{n, 1} \lt q_n \le r_{n, 2}\). But when we have (possibly uncountably) infinite number of pairs, \(\{(r_{j, 1}, r_{j, 2}) \vert j \in J\}\), there can be an objection against our just claiming that we can choose a \(q_j \in \mathbb{Q}\) such that \(r_{j, 1} \lt q_j \le r_{j, 2}\) for each \(j \in J\). The axiom of choice could be employed, but let us do without it. If there is a systematic way of choosing a \(q\) for any \((r_1, r_2)\), we can claim that we can choose a \(q_j \in \mathbb{Q}\) such that \(r_{j, 1} \lt q_j \le r_{j, 2}\) for each \(j \in J\) by that way.
Of course, there can be many other ways, but presenting a way is enough for our purpose of claiming that we can choose a \(q_j \in \mathbb{Q}\) for each \(j \in J\).
\(q\) can be also systematically chosen satisfying \(q \in \mathbb{Q} \land r_1 \lt q \lt r_2\), because we can take \((r_1 + r_2) / 2\) and take \(r_1 \lt q \le (r_1 + r_2) / 2 \lt r_2\), applying this proposition.
3: Proof
Whole Strategy: Step 1: let \(r_1\) and \(r_2\) be expressed as the decimals without any trailing '999...'; Step 2: when \(0 \le r_2\), take \(q\) as is mentioned in Statements, and see that \(q \in \mathbb{Q}\) and \(r_1 \lt q \le r_2\); Step 3: when \(r_2 \lt 0\), take \(q\) as is mentioned in Statements, and see that \(q \in \mathbb{Q}\) and \(r_1 \lt q \le r_2\); Step 4: conclude the proposition.
Step 1:
Let \(r_1\) and \(r_2\) be expressed as the decimals without any trailing '999...' (for example, '12.3' instead of '12.2999...'), which makes the expressions unique. When a decimal is finite, let the decimal have the trailing '000...'.
Step 2:
Let us suppose that \(0 \le r_2\).
\(r_1 \lt 0\) or \(0 \le r_1\).
When \(r_1 \lt 0\), let \(q = 0\).
Then, \(q \in \mathbb{Q}\) and \(r_0 \lt q \le r_2\).
Let us suppose that \(0 \le r_1\).
There is the 1st digit on which \(r_1\) and \(r_2\) disagree.
Let \(q\) be \(r_2\) with the subsequent digits cut off.
Then, \(q \in \mathbb{Q}\), because it has the finite decimal, and \(r_1 \lt q \le r_2\), obviously.
An example is \(r_1 = 12.344678..., r_2 = 12.345678...\), and \(q = 12.345\).
Step 3:
Let us suppose that \(r_2 \lt 0\).
There is the 1st digit on which \(r_1\) and \(r_2\) disagree.
There is the 1st digit of \(r_2\) after that that is not '9' (because the expression is without any trailing '999...').
Let \(q\) be \(r_2\) with the digit incremented by \(1\) and the subsequent digits cut off.
Then, \(q \in \mathbb{Q}\), because it has the finite decimal, and \(r_1 \lt q \le r_2\), because as the disagreeing digit is not changed, \(r_1 \lt q\) holds.
An example is \(r_1 = - 12.346678..., r_2 = - 12.3459678...\), and \(q = - 12.34597\).
Step 4:
\(q\) has been chosen without any arbitrariness satisfying \(q \in \mathbb{Q} \land r_1 \lt q \le r_2\).
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 for (countably) compact metric space with induced topology, non-convergent sequence has more than \(1\) points to which subsequences converge
Topics
About:
metric 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 (countably) compact metric space with the induced topology, any non-convergent sequence has some more than \(1\) points to which some subsequences converge.
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:
\(M\): \(\in \{\text{ the countably compact metric spaces }\}\), with the induced topology
\(s\): \(: J \to M\), \(\in \text{ the sequences }\)
//
Statements:
\(s \notin \{\text{ the convergent sequences }\}\)
\(\implies\)
\(\exists s^`, \widetilde{s^`} \in \{\text{ the convergent subsequences of } s\} (lim s^` \neq lim \widetilde{s^`})\)
//
If \(M\) is compact, \(M\) is countably compact, by the proposition that any metric space is compact if and only if it is countably compact, so, \(M\) can be required to be compact.
2: Note
There may not be some more than \(2\) points to which some subsequences converge.
For example, let \(M = [-1, 1]\) and \(s: \mathbb{N} \to [-1, 1], n \mapsto - 1 / 2 \text{ when } n \text{ is even }; \mapsto 1 / 2 \text{ when } n \text{ is odd }\), then, \(s\) is not convergent, and there are a subsequence that converges to \(- 1 / 2\) and a subsequence that converges to \(1 / 2\), but there is no other point to which a subsequence converges.
3: Proof
Whole Strategy: Step 1: see that \(M\) is compact; Step 2: see that \(M\) is sequentially compact; Step 3: take any convergent subsequence of \(s\), \(s^`\), such that \(lim s^` = m\); Step 4: for each \(\epsilon\), take a finite open cover of \(M\), \(\{B_{m, \epsilon}\} \cup \{B_{m_j, \epsilon / 2} \vert j \in J\}\), and see that for an \(\epsilon\), a \(B_{m_j, \epsilon / 2}\) contains some infinite points of \(s\), which determines the subsequence of \(s\), \(\widetilde{s^`}'\); Step 5: take a convergent subsequence of \(\widetilde{s^`}'\), \(\widetilde{s^`}\).
Step 1:
\(M\) is compact, by the proposition that any metric space is compact if and only if it is countably compact.
In fact, \(M\) can be required to be compact, because then, \(M\) is countably compact.
Step 2:
\(M\) is sequentially compact, by the proposition that any metric space with the induced topology is 1st-countable and the proposition that any 1st-countable topological space is sequentially compact if the space is countably compact.
Step 3:
There is a convergent subsequence of \(s\), \(s^`: J^` \to M\), with \(m := lim s^`\), because \(M\) is sequentially compact: refer to the proposition that for any metric space, if and only if each sequence on it has a convergent subsequence, each sequence on it from the natural numbers set has a convergent subsequence.
Step 4:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let us take the open cover of \(M\), \(\{B_{m, \epsilon}\} \cup \{B_{m_j, \epsilon / 2} \vert m_j \in M \setminus B_{m, \epsilon}\}\).
That is indeed an open cover, because for each \(m' \in M\), \(m' \in B_{m, \epsilon}\) or \(m' \in M \setminus B_{m, \epsilon}\), but when \(m' \in M \setminus B_{m, \epsilon}\), \(m' \in B_{m_j, \epsilon / 2}\) where \(m_j = m'\).
As \(M\) is compact, there is a finite subcover, \(\{B_{m, \epsilon}\} \cup \{B_{m_j, \epsilon / 2} \vert j \in J\}\), where \(J\) is a finite index set.
If for each \(\epsilon\), each element of \(\{B_{m_j, \epsilon / 2} \vert j \in J\}\) contained only some finite points of \(s\), there would be an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(s (J_n) \in B_{m, \epsilon}\), which would mean that \(dist (m, s (J_n)) \lt \epsilon\), which would mean that \(s\) converged to \(m\), a contradiction against that \(s\) was not convergent.
So, there is an \(\epsilon\) such that a \(B_{m_j, \epsilon / 2}\) contains some infinite points of \(s\).
That determines the subsequence of \(s\), \(\widetilde{s^`}': {J^`}' \to M = s \circ f'\), where \({J^`}' = \{j^` \in J \vert s (j^`) \in B_{m_j, \epsilon / 2}\}\) and \(f': {J^`}' \to J, j^` \to j^`\), which is indeed a subsequence of \(s\), because for each \(j^`_1, j^`_2 \in {J^`}'\) such that \(j^`_1 \lt j^`_2\), \(f' (j^`_1) = j^`_1 \lt j^`_2 = f' (j^`_2)\), and for each \(j \in J\), there is a \(j^` \in {J^`}'\) such that \(j \le j^` = f' (j^`)\), because \({J^`}'\) is infinite.
Step 5:
There is a convergent subsequence of \(\widetilde{s^`}'\), \(\widetilde{s^`}: J^` \to M = \widetilde{s^`}' \circ f\), with \(lim \widetilde{s^`} = \widetilde{m}\), because \(M\) is sequentially compact.
\(\widetilde{s^`} = \widetilde{s^`}' \circ f = s \circ f' \circ f\) is a subsequence of \(s\), because for each \(j^`_1, j^`_2 \in J^`\) such that \(j^`_1 \lt j^`_2\), \(f' \circ f (j^`_1) \lt f' \circ f (j^`_2)\), because \(f (j^`_1) \lt f (j^`_2)\), so, \(f' \circ f (j^`_1) \lt f' \circ f (j^`_2)\), and for each \(j \in J\), there is a \({j^`}' \in {J^`}'\) such that \(j \le f' ({j^`}')\) and there is a \(j^` \in J^`\) such that \({j^`}' \le f (j^`)\), so, \(j \le f' ({j^`}') \le f' \circ f (j^`)\).
\(dist (\widetilde{m}, m_j) \le \epsilon / 2\), because for each \(\epsilon' \in \mathbb{R}\) such that \(0 \lt \epsilon'\), there is an \(n \in \mathbb{N} \setminus \{0\}\) such that \(dist (\widetilde{m}, \widetilde{s^`} ({J^`}_n)) \lt \epsilon'\), because \(\widetilde{s^`}\) converges to \(\widetilde{m}\), and \(dist (\widetilde{m}, m_j) \le dist (\widetilde{m}, \widetilde{s^`} ({J^`}_n)) + dist (\widetilde{s^`} ({J^`}_n), m_j) \lt \epsilon' + \epsilon / 2\), so, \(dist (\widetilde{m}, m_j) \le \epsilon / 2\), by the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
\(\epsilon \lt dist (m, m_j) \le dist (m, \widetilde{m}) + dist (\widetilde{m}, m_j) \le dist (m, \widetilde{m}) + \epsilon / 2\), so, \(\epsilon / 2 = \epsilon - \epsilon / 2 \lt dist (m, \widetilde{m})\).
So, \(m \neq \widetilde{m}\).
So, there are at least some \(2\) subsequences, \(s^`\) and \(\widetilde{s^`}\) such that \(lim s^` \neq lim \widetilde{s^`}\).
References
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