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description/proof of that for finite set of points on real vectors space, if for point, set of subtractions of point from other points are linearly independent, it is so for each point
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definition of linearly independent subset of module
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of linearly independent subset of module.
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:
\( R\): \(\in \{\text{ the rings }\}\)
\( M\): \(\in \{\text{ the } R \text{ modules }\}\)
\(*S\): \(\subseteq M\), \(\in \{\text{ the possibly uncountable sets }\}\)
//
Conditions:
\(\forall S' \subseteq S, S' \in \{\text{ the finite sets }\}\)
(
\(\sum_{p_j \in S'} r^j p_j = 0, r^j \in R\)
\(\implies\)
\(\forall j (r^j = 0)\)
)
//
2: Natural Language Description
For any module, \(M\), over any ring, \(R\), any (possibly uncountable) subset, \(S \subseteq M\), such that for each finite subset, \(S' \subseteq S\), \(\sum_{p_j \in S'} r^j p_j = 0\) implies that \(r^j = 0\) for each \(j\), where \(r^j\) is any element of \(R\)
3: Note
As any vectors space is a module, 'linearly independent subset of vectors space' is nothing but 'linearly independent subset of module'.
We need to think of the linear combination of each finite subset instead of the linear combination of possibly infinite \(S\), because the convergence of the linear combination of any infinite elements is not defined without any extra structure like topology or metric.
References
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definition of %ring name% module
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of %ring name% module.
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:
\( R\): \(\in \{\text{ the rings }\}\)
\(*M\): \(\in \{\text{ the sets }\}\) with any \(+: M \times M \to M\) (addition) operation and any \(.: R \times M \to M\) (scalar multiplication) operation
//
Conditions:
1) \(\forall m_1, m_2 \in M (m_1 + m_2 \in M)\) (closed-ness under addition)
\(\land\)
2) \(\forall m_1, m_2 \in M (m_1 + m_2 = m_2 + m_1)\) (commutativity of addition)
\(\land\)
3) \(\forall m_1, m_2, m_3 \in M ((m_1 + m_2) + m_3 = m_1 + (m_2 + m_3))\) (associativity of additions)
\(\land\)
4) \(\exists 0 \in M (\forall m \in M (m + 0 = m))\) (existence of 0 element)
\(\land\)
5) \(\forall m \in M (\exists m' \in M (m' + m = 0))\) (existence of inverse vector)
\(\land\)
6) \(\forall m \in M, \forall r \in R (r . m \in M)\) (closed-ness under scalar multiplication)
\(\land\)
7) \(\forall m \in M, \forall r_1, r_2 \in R ((r_1 + r_2) . m = r_1 . m + r_2 . m)\) (scalar multiplication distributability for scalars addition)
\(\land\)
8) \(\forall m_1, m_2 \in M, \forall r \in R (r . (m_1 + m_2) = r . m_1 + r . m_2)\) (scalar multiplication distributability for elements addition)
\(\land\)
9) \(\forall m \in M, \forall r_1, r_2 \in R ((r_1 r_2) . m = r_1 . (r_2 . m))\) (associativity of scalar multiplications)
\(\land\)
10) \(\forall m \in M (1 . m = m)\) (identity of 1 multiplication)
//
2: Natural Language Description
Any set, \(M\), with any \(+: M \times M \to M\) (addition) operation and any \(.: R \times M \to M\) (scaler multiplication) operation with respect to any ring, \(R\), that satisfies these conditions: 1) for any elements, \(m_1, m_2 \in M\), \(m_1 + m_2 \in M\) (closed-ness under addition); 2) for any elements, \(m_1, m_2 \in M\), \(m_1 + m_2 = m_2 + m_1\) (commutativity of addition); 3) for any elements, \(m_1, m_2, m_3 \in M\), \((m_1 + m_2) + m_3 = m_1 + (m_2 + m_3)\) (associativity of additions); 4) there is a 0 element, \(0 \in M\), such that for each \(m \in M\), \(m + 0 = m\) (existence of 0 element); 5) for any element, \(m \in M\), there is an inverse element, \(m' \in M\), such that \(m' + m = 0\) (existence of inverse vector); 6) for any element, \(m \in M\), and any scalar, \(r \in R\), \(r . m \in M\) (closed-ness under scalar multiplication); 7) for any element, \(m \in M\), and any scalars, \(r_1, r_2 \in R\), \((r_1 + r_2) . m = r_1 . m + r_2 . m\) (scalar multiplication distributability for scalars addition); 8) for any elements, \(m_1, m_2 \in M\), and any scalar, \(r \in R\), \(r . (m_1 + m_2) = r . m_1 + r . m_2\) (scalar multiplication distributability for elements addition); 9) for any element, \(m \in M\), and any scalars, \(r_1, r_2 \in R\), \((r_1 r_2) . m = r_1 . (r_2 . m)\) (associativity of scalar multiplications); 10) for any element, \(m \in M\), \(1 . m = m\) (identity of 1 multiplication)
3: Note
\(.\) is often omitted in notations like \(r m\) instead of \(r . m\).
The requirements 1) ~ 10) are in fact parallel to those for vectors space; the difference between vectors space and module is only that the scalar structure is a field or a ring, which makes a significant difference, because for example, for a module, \(r^1 m_1 + r^2 m_2 + r^3 m_3 = 0\) where \(r^1 \neq 0\) does not imply that \(m_1\) is a linear combination of \(m_2\) and \(m_3\), because \({r^1}^{-1}\) is not guaranteed to exist.
References
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definition of product map
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of product map.
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 1
Here is the rules of Structured Description.
Entities:
\( A\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\( \{S_\alpha\}\): \(\alpha \in A\), \(S_\alpha \in \{\text{ the sets }\}\)
\( \{S'_\alpha\}\): \(\alpha \in A\), \(S'_\alpha \in \{\text{ the sets }\}\)
\( \{f_\alpha\}\): \(\alpha \in A\), \(: S_\alpha \to S'_\alpha\)
\(*\times_{\alpha \in A} f_\alpha\): \(:\times_{\alpha \in A} S_\alpha \to \times_{\alpha \in A} S'_\alpha, (\alpha \mapsto f (\alpha)) \mapsto (\alpha \mapsto f_\alpha (f (\alpha)))\)
//
Conditions:
//
2: Natural Language Description 1
For any possibly uncountable index set, \(A\), any sets, \(\{S_\alpha \vert \alpha \in A\}\), any sets, \(\{S'_\alpha \vert \alpha \in A\}\), and any maps, \(\{f_\alpha: S_\alpha \to S'_\alpha\}\), \(\times_{\alpha \in A} f_\alpha :\times_{\alpha \in A} S_\alpha \to \times_{\alpha \in A} S'_\alpha\), \((\alpha \mapsto f (\alpha)) \mapsto (f': \alpha \mapsto f_\alpha (f (\alpha)))\)
3: Structured Description 2
Here is the rules of Structured Description.
Entities:
\( J\): \(= \{1, ..., n\}\)
\( \{S_j\}\): \(j \in J\), \(S_j \in \{\text{ the sets }\}\)
\( \{S'_j\}\): \(j \in J\), \(S'_j \in \{\text{ the sets }\}\)
\( \{f_j\}\): \(j \in J\), \(: S_j \to S'_j\)
\(*f_1 \times f_2 \times ... \times f_n\): \(: S_1 \times S_2 \times ... \times S_n \to S'_1 \times S'_2 \times ... \times S'_n, (p_1, p_2, ..., p_n) \mapsto (f_1 (p_1), f_2 (p_2), ..., f_n (p_n))\)
Conditions:
//
4: Natural Language Description 2
For any finite number of sets, \(S_1, S_2, ..., S_n\), any same number of sets, \(S'_1, S'_2, ..., S'_n\), and any same number of maps, \(f_1: S_1 \to S'_1, f_2: S_2 \to S'_2, ..., f_n: S_n \to S'_n\), \(f_1 \times f_2 \times ... \times f_n: S_1 \times S_2 \times ... \times S_n \to S'_1 \times S'_2 \times ... \times S'_n\), \((p_1, p_2, ..., p_n) \mapsto (f_1 (p_1), f_2 (p_2), ..., f_n (p_n))\)
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description/proof of sufficient conditions for existence of unique global solution on interval for Euclidean-normed Euclidean vectors space ODE
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description/proof of that restriction of continuous embedding on domain and codomain is continuous embedding
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 restriction of any continuous embedding on the domain and the codomain is a continuous embedding.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous embeddings }\}\)
\(S_1\): \(\subseteq T_1\)
\(S_2\): \(\subseteq T_2\), \(f (S_1) \subseteq S_2\)
\(f'\): \(: S_1 \to S_2, p \mapsto f (p)\)
//
Statements:
\(f' \in \{\text{ the continuous embeddings }\}\)
//
2: Natural Language Description
For any topological spaces, \(T_1, T_2\), any continuous embedding, \(f: T_1 \to T_2\), any subset, \(S_1 \subseteq T_1\), and any subset, \(S_2 \subseteq T_2\) such that \(f (S_1) \subseteq S_2\), \(f': S_1 \to S_2, p \mapsto f (p)\) is a continuous embedding.
3: Proof
\(f'\) is injective, because \(f\) is so.
\(f'\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Let us denote the codomain restriction of \(f'\) as \(f'' : S_1 \to f' (S_1)\).
\(f''\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
As \(f''\) is bijective, there is the inverse, \(f''^{-1}: f' (S_1) \to S_1\). Is \(f''^{-1}\) continuous?
For any open subset, \(U \subseteq S_1\), is \({f''^{-1}}^{-1} (U) = f'' (U)\) open on \(f' (S_1)\)?
\(U = U' \cap S_1\) for an open subset, \(U' \subseteq T_1\). \(f'' (U) = f'' (U' \cap S_1) = f'' (U') \cap f'' (S_1)\), by the proposition that for any injective map, the map image of the intersection of any sets is the intersection of the map images of the sets, \(= f (U') \cap f'' (S_1) = U'' \cap f (T_1) \cap f'' (S_1)\) where \(U'' \subseteq T_2\) is an open subset, because \(f: T_1 \to f (T_1)\) is a homeomorphism (so, \(f (U')\) is open on \(f (T_1)\)), \(= U'' \cap f'' (S_1)\), which is open on \(f'' (S_1) = f' (S_1)\).
So, yes, \(f''^{-1}\) is continuous, and \(f''\) is a homeomorphism.
References
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description/proof of that for infinite product topological space and closed subset, point on product space whose each finite-components-projection belongs to corresponding projection of subset belongs to 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 infinite product topological space and any closed subset, any point on the product space whose each finite-components-projection belongs to the corresponding projection of the subset belongs to 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:
\(A\): \(\in \{\text{ the possibly uncountable infinite index sets }\}\)
\(\{T_\alpha \vert \alpha \in A\}\): \(T_\alpha \in \{\text{ the topological spaces }\}\)
\(T\): \(= \times_{\alpha \in A} T_\alpha\) with the product topology
\(C\): \(\subseteq T\), \(\in \{\text{ the closed subsets }\}\)
\(p\): \(\in T\)
//
Statements:
\(\forall J \subset A, J \in \{\text{ the finite index sets }\} (\pi_{J} (p) \in \pi_{J} (C))\), where \(\pi_{J}: T \to \times_{j \in J} T_j\) is the projection
\(\implies\)
\(p \in C\)
//
2: Natural Language Description
For any possibly uncountable infinite index set, \(A\), any topological spaces, \(\{T_\alpha \vert \alpha \in A\}\), the product topological space, \(T := \times_{\alpha \in A} T_\alpha\), any closed \(C \subseteq T\), and any \(p \in T\), \(\pi_{J} (p) \in \pi_{J} (C)\) for each \(J \subset A\), where \(J\) is a finite index set and \(\pi_{J}: T \to \times_{j \in J} T_j\) is the projection, implies that \(p \in C\).
3: Note 1
A typical case is \(T = T_1 \times T_2 \times ...\), a infinitely countable product, and \(\pi_{J} (p) \in \pi_{J} (C)\) where \(J = \{1, ..., k\}\) for each \(k\) implies \(p \in C\): the requirement for such \(J\)s implies the requirement for each \(J' \subseteq J\), so, implies the requirement for each finite \(J' \subseteq A\).
4: Proof
Let us suppose that \(\pi_{J} (p) \in \pi_{J} (C)\) for each \(J\).
\(T \setminus C \subseteq T\) is open.
Let us suppose that \(p \notin C\).
\(p \in T \setminus C\). There would be an open neighborhood, \(U_p \subseteq T \setminus C\), of \(p\). \(U_p = \cup_{\beta \in B} \times_{\alpha \in A} U_{\beta, \alpha}\), where \(B\) would be a possibly uncountable index set and \(U_{\beta, \alpha} \subseteq T_\alpha\) would be open while only finite number of \(U_{\beta, \alpha}\) s would not be \(T_\alpha\) s for each \(\beta\) (see Note of the definition of product topology).
There would be a \(\beta\) such that \(p \in \times_{\alpha \in A} U_{\beta, \alpha} \subseteq T \setminus C\). As only finite of \(U_{\beta, \alpha}\) s would not be \(T_\alpha\) s, let \(J\) be of the finite components. \(\pi_J (p) \in \pi_J (\times_{\alpha \in A} U_{\beta, \alpha}) = \times_{j \in J} U_{\beta, j}\). Then, there would be a \(p' \in C\) such that \(\pi_J (p') = \pi_J (p) \in \times_{j \in J} U_{\beta, j}\). But \(p' \in \times_{\alpha \in A} U_{\beta, \alpha}\), because \(U_{\beta, \alpha} = T_\alpha\) for each \(\alpha \notin J\) and so, \(p'^\alpha \in U_{\beta, \alpha}\) when \(\alpha \in J\) and when \(\alpha \in A \setminus J\).
So, \(p' \in \times_{\alpha \in A} U_{\beta, \alpha} \subseteq \cup_{\beta \in B} \times_{\alpha \in A} U_{\beta, \alpha} = U_p \subseteq T \setminus C\), which would mean \(p' \notin C\), a contradiction.
So, \(p \in C\).
5: Note 2
When \(C \subseteq T\) is not closed, \(p \in C\) is not necessarily implied as is proved in another article.
References
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description/proof of that for infinite product topological space and subset, point on product space whose each finite-components-projection belongs to corresponding projection of subset does not necessarily belong to 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 an infinite product topological space and a subset, a point on the product space whose each finite-components-projection belongs to the corresponding projection of the subset does not necessarily belong to 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:
\(A\): \(\in \{\text{ the possibly uncountable infinite index sets }\}\)
\(\{T_\alpha \vert \alpha \in A\}\): \(T_\alpha \in \{\text{ the topological spaces }\}\)
\(T\): \(= \times_{\alpha \in A} T_\alpha\) with the product topology
\(S\): \(\subseteq T\)
\(p\): \(\in T\)
//
Statements:
\(\forall J \subset A, J \in \{\text{ the finite index sets }\} (\pi_{J} (p) \in \pi_{J} (S))\) does not necessarily imply \(p \in S\), where \(\pi_{J}: T \to \times_{j \in J} T_j\) is the projection.
//
2: Natural Language Description
For any possibly uncountable infinite index set, \(A\), any topological spaces, \(\{T_\alpha \vert \alpha \in A\}\), the product topological space, \(T := \times_{\alpha \in A} T_\alpha\), any \(S \subseteq T\), and any \(p \in T\), \(\pi_{J} (p) \in \pi_{J} (S)\) for each \(J \subset A\), where \(J\) is a finite index set and \(\pi_{J}: T \to \times_{j \in J} T_j\) is the projection, does not necessarily imply that \(p \in S\).
3: Note 1
A typical case is \(T = T_1 \times T_2 \times ...\), a infinitely countable product, and \(\pi_{J} (p) \in \pi_{J} (S)\) where \(J = \{1, ..., k\}\) for each \(k\) does not imply \(p \in S\).
4: Proof
A counterexample suffices.
Let each \(T_\alpha\) have at least 2 points, \(p_{\alpha, 1}, p_{\alpha, 2} \in T_\alpha\), let \(S = T \setminus \times_{\alpha \in A} \{p_{\alpha, 1}\}\), and let each \(\alpha\) component of \(p\) be \(p_{\alpha, 1}\).
\(p \in T\). \(\pi_{J} (p) \in \pi_{J} (S)\) for each \(J\), because \(S\) has at least 1 point, \(p'\), whose \(J\) components are \(p_{j, 1}\) s (\(j \in J\)) but whose \(\alpha\) component is \(p_{\alpha, 2}\) where \(\alpha \notin J\); \(p' \in S\), because \(p_{\alpha, 2} \notin \{p_{\alpha, 1}\}\).
But \(p \notin S\), because \(p \in \times_{\alpha \in A} \{p_{\alpha, 1}\}\).
Letting \(A\) be countable does not help. Let \(A = \{1, 2, ...\}\), let each \(T_j\) have at least 2 points, \(p_{j, 1}, p_{j, 2} \in T_j\), let \(S = T \setminus (\{p_{1, 1}\} \times \{p_{2, 1}\} \times ...)\), and let \(p\) be \((p_{1, 1}, p_{2, 1}, ...)\).
\(p \in T\). \(\pi_{J} (p) \in \pi_{J} (S))\) for each \(J\), because, for example, when \(J = \{2, 4\}\), \((p_{2, 2}, p_{2, 1}, p_{3, 1}, p_{4, 1}, ...), (p_{1, 1}, p_{2, 1}, p_{3, 2}, p_{4, 1}, ...), ... \in S\), while \(\pi_{J} (p) = \pi_{J} ((p_{2, 2}, p_{2, 1}, p_{3, 1}, p_{4, 1}, ...))) = (p_{2, 1}, p_{4, 1})\).
But \(p \notin S\), because \(p \in \{p_{1, 1}\} \times \{p_{2, 1}\} \times ...\).
5: Note 2
When \(S \subseteq T\) is closed, \(p \in S\) is implied as is proved in another article.
References
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