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description/proof of that for \(C^\infty\) immersion between \(C^\infty\) manifolds with boundary, its global differential is \(C^\infty\) immersion
Topics
About:
\(C^\infty\) manifold
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 \(C^\infty\) immersion between any \(C^\infty\) manifolds with boundary, its global differential is a \(C^\infty\) immersion.
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:
\(M_1\): \(\in \{\text{ the } d_1 \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\(M_2\): \(\in \{\text{ the } d_2 \text{ -dimensional } C^\infty \text{ manifolds with boundary }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the } C^\infty \text{ immersions }\}\)
\(d f\): \(: T M_1 \to T M_2\), \(= \text{ the global differential }\)
//
Statements:
\(d f \in \{\text{ the } C^\infty \text{ immersions }\}\).
//
2: Proof
Whole Strategy: do it in the 2 parts: the 1st part: \(M_2\) has the empty boundary; the 2nd part: \(M_2\) has a nonempty boundary; Step 1: suppose that \(M_2\) has the empty boundary; Step 2: around each \(m \in M_1\), take a chart, \((U_m \subseteq M_1, \phi_m)\), and a chart, \((U_{f (m)} \subseteq M_2, \phi_{f (m)})\), such that \(\widehat{f} := \phi_{f (m)} \circ f \circ {\phi_m}^{-1}\) is \(: (x^1, ..., x^{d_1}) \mapsto (x^1, ..., x^{d_1}, 0, ..., 0)\); Step 3: let \(\pi_1: T M_1 \to M_1\) and \(\pi_2: T M_2 \to M_2\) be the projections; take the induced charts, \(({\pi_1}^{-1} (U_m) \subseteq T M_1, \widetilde{\phi_m})\) and \(({\pi_2}^{-1} (U_{f (m)}) \subseteq T M_2, \widetilde{\phi_{f (m)}})\), and see that \(\widehat{d f} := \widetilde{\phi_{f (m)}} \circ d f \circ {\widetilde{\phi_m}}^{-1}\) is \(: (v^1, ..., v^k, x^1, ..., x^{d_1}) \mapsto (v^1, ..., v^k, 0, ..., 0, x^1, ..., x^{d_1}, 0, ..., 0)\); Step 4: see that \(d f\) is a \(C^\infty\) immersion; Step 5: suppose that \(M_2\) has a nonempty boundary; Step 6: take the double of \(M_2\), \(D (M_2)\), let \(\widetilde{M_2} \subseteq D (M_2)\) be a regular domain diffeomorphic to \(M_2\), with a diffeomorphism, \(g: M_2 \to \widetilde{M_2}\), let \(\widetilde{\iota}: \widetilde{M_2} \to D (M_2)\) be the inclusion, take \(h := \widetilde{\iota} \circ g \circ f\), apply the 1st part conclusion to see that \(d h\) is a \(C^\infty\) immersion, and see that \(d f\) is a \(C^\infty\) immersion.
Step 1:
Le us suppose that \(M_2\) has the empty boundary.
Step 2:
Around each \(m \in M_1\), let us take a chart, \((U_m \subseteq M_1, \phi_m)\), and a chart, \((U_{f (m)} \subseteq M_2, \phi_{f (m)})\), such that \(f (U_m) \subseteq U_{f (m)}\) and \(\widehat{f} := \phi_{f (m)} \circ f \circ {\phi_m}^{-1}: \phi_m (U_m) \to \phi_{f (m)} (U_{f (m)})\), the coordinates function of \(f\), is \(: (x^1, ..., x^{d_1}) \mapsto (x^1, ..., x^{d_1}, 0, ..., 0)\), which is possible by the rank theorem for \(C^\infty\) immersion (which requires \(M_2\) to be without boundary, which is the reason why we have made the supposition of Step 1).
Step 3:
Let \(\pi_1: T M_1 \to M_1\) and \(\pi_2: T M_2 \to M_2\) be the projections.
Let us take the induced charts, \(({\pi_1}^{-1} (U_m) \subseteq T M_1, \widetilde{\phi_m})\) and \(({\pi_2}^{-1} (U_{f (m)}) \subseteq T M_2, \widetilde{\phi_{f (m)}})\).
It is a well-known fact that \(\widehat{d f} := \widetilde{\phi_{f (m)}} \circ d f \circ {\widetilde{\phi_m}}^{-1}: \widetilde{\phi_m} ({\pi_1}^{-1} (U_m)) \to \widetilde{\phi_{f (m)}} ({\pi_2}^{-1} (U_{f (m)}))\) is \(: (v^1, ..., v^{d_1}, x^1, ..., x^{d_1}) \mapsto (v^1, ..., v^{d_1}, 0, ..., 0, x^1, ..., x^{d_1}, 0, ..., 0)\): the components, \((v^1, ..., v^{d_1})\), are mapped to \((\partial_j \widehat{f}^1 v^j, ..., \partial_j \widehat{f}^{d_2} v^j)\), but \(\widehat{f}^j = x^j\) for each \(1 \le j \le d_1\) and \(\widehat{f}^j = 0\) for each \(d_1 + 1 \le j \le d_2\).
So, \(d f\) is \(C^\infty\).
Step 4:
Let \(\pi'_1: T T M_1 \to T M_1\) and \(\pi'_2: T T M_2 \to T M_2\) be the projections.
There are the induced charts, \(({\pi'_1}^{-1} ({\pi_1}^{-1} (U_m)) \subseteq T T M_1, \widetilde{\widetilde{\phi_m}})\) and \(({\pi'_2}^{-1} ({\pi_2}^{-1} (U_{f (m)})) \subseteq T T M_2, \widetilde{\widetilde{\phi_{f (m)}}})\).
It is a well-known fact that \(\widehat{d d f} := \widetilde{\widetilde{\phi_{f (m)}}} \circ d d f \circ {\widetilde{\widetilde{\phi_m}}}^{-1}: \widetilde{\widetilde{\phi_m}} ({\pi'_1}^{-1} ({\pi_1}^{-1} (U_m))) \to \widetilde{\widetilde{\phi_{f (m)}}} ({\pi'_2}^{-1} ({\pi_2}^{-1} (U_{f (m)})))\) is \(: (w^1, ..., w^{2 d_1}, v^1, ..., v^{d_1}, x^1, ..., x^{d_1}) \mapsto (w^1, ..., w^{d_1}, 0, ..., 0, w^{d_1 + 1}, ..., w^{2 d_1}, 0, ..., 0, v^1, ..., v^{d_1}, 0, ..., 0, x^1, ..., x^{d_1}, 0, ..., 0)\): the components, \((w^1, ..., w^{2 d_1})\), are mapped to \((\partial_j \widehat{d f}^1 w^j, ..., \partial_j \widehat{d f}^{2 d_2} w^j)\) where \(\partial_j\) s for \(1 \le j \le d_1\) are by \(v^j\) s and \(\partial_j\) s for \(d_1 + 1 \le j \le 2 d_1\) are by \(x^{j - d_1}\) s, but \(\widehat{d f}^j = v^j\) for each \(1 \le j \le d_1\), \(\widehat{d f}^j = 0\) for each \(d_1 + 1 \le j \le d_2\), \(\widehat{d f}^j = x^{j - d_2}\) for each \(d_2 + 1 \le j \le d_2 + d_1\), and \(\widehat{d f}^j = 0\) for each \(d_2 + d_1 + 1 \le j \le 2 d_2\).
That is obviously injective.
So, \(d f\) is a \(C^\infty\) immersion.
Step 5:
Let us suppose that \(M_2\) has a nonempty boundary.
Step 6:
Let us take the double of \(M_2\), \(D (M_2)\).
\(D (M_2)\) is a \(C^\infty\) manifold without boundary that has a regular domain, \(\widetilde{M_2}\), that is diffeomorphic to \(M_2\).
Let a diffeomorphism be \(g: M_2 \to \widetilde{M_2}\).
Let \(\widetilde{\iota}: \widetilde{M_2} \to D (M_2)\) be the inclusion, which is a \(C^\infty\) immersion (in fact, a \(C^\infty\) embedding), because \(\widetilde{M_2}\) is an embedded submanifold (in fact, a regular domain) of \(D (M_2)\).
Let us think of \(h: M_1 \to D (M_2) = \widetilde{\iota} \circ g \circ f: M_1 \to M_2 \to \widetilde{M_2} \to D (M_2)\).
\(h\) is \(C^\infty\), by the proposition that for any maps between any arbitrary subsets of any \(C^\infty\) manifolds with boundary \(C^k\) at corresponding points, where \(k\) includes \(\infty\), the composition is \(C^k\) at the point.
\(h\) is a \(C^\infty\) immersion, because \(d h = d \widetilde{\iota} \circ d g \circ d f\) and \(d f\), \(d g\), and \(d \widetilde{\iota}\) are injective.
By the 1st part conclusion, \(d h = d \widetilde{\iota} \circ d g \circ d f\) is a \(C^\infty\) immersion, which means that \(d d h = d d \widetilde{\iota} \circ d d g \circ d d f\) is injective on each fiber.
Then, \(d d f\) is injective on each fiber, because otherwise, the 2 vectors that were mapped to the same vector under \(d d f\) would not be mapped to any different vectors under \(d d h\).
So, \(d f\) is a \(C^\infty\) immersion.
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 global differential of \(C^\infty\) map between \(C^\infty\) manifolds with boundary
Topics
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of global differential of \(C^\infty\) map between \(C^\infty\) manifolds with boundary.
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:
\( M_1\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\( M_2\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\( f\): \(: M_1 \to M_2\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
\( (T M_1, M_1, \pi_1)\): \(= \text{ the tangent vectors bundle }\)
\( (T M_2, M_2, \pi_2)\): \(= \text{ the tangent vectors bundle }\)
\(*d f\): \(: T M_1 \to T M_2, v \mapsto d f_{\pi_1 (v)} v\)
//
Conditions:
//
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 measurable subspace
Topics
About:
measurable space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of measurable 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:
\( (M', A')\): \(\in \{\text{ the measurable spaces }\}\)
\( M\): \(\subseteq M'\)
\( A\): \(= \text{ the subspace } \sigma\text{ -algebra of the subset, } M \text{, of } (M', A')\)
\(*(M, A)\): \(\in \{\text{ the measurable spaces }\}\)
//
Conditions:
//
2: Natural Language Description
For any measurable space, \((M', A')\), any subset, \(M \subseteq M'\), and the subspace \(\sigma\)-algebra of the subset, \(M\), of \((M', A')\), \(A\), the measurable space, \((M, A)\)
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 subspace \(\sigma\)-algebra of subset of measurable space
Topics
About:
measurable space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of subspace \(\sigma\)-algebra of subset of measurable 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:
\( (M', A')\): \(\in \{\text{ the measurable spaces }\}\)
\( M\): \(\subseteq M'\)
\(*A\): \(= \{W' \cap M \vert W' \in A'\}\)
//
Conditions:
//
2: Natural Language Description
For any measurable space, \((M', A')\), and any subset, \(M \subseteq M'\), \(A: = \{W' \cap M \vert W' \in A'\}\)
3: Note
\(A\) is indeed a \(\sigma\)-algebra of \(M\): 1) \(M = M' \cap M \in A\), because \(M' \in A'\); 2) for each \(W \in A\), \(M \setminus W \in A\), because \(M \setminus W = M \setminus (W' \cap M) = (M' \setminus W') \cap M\), by the proposition that the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd set, and \(M' \setminus W' \in A'\); 3) for each infinite sequence, \(W_1, W_2, ...\), where \(W_j \in A\), \(\cup_j W_j \in A\), because \(\cup_j W_j = \cup_j (W'_j \cap M) = (\cup_j W'_j) \cap M\), 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, and \(\cup_j W'_j \in A'\).
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 set minus (set minus set) is not necessarily but contains (1st set minus 2nd set) minus 3rd 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 any set minus (any set minus any set) is not necessarily equal to but contains (the 1st set minus the 2nd set) minus the 3rd 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(S_3\): \(\in \{\text{ the sets }\}\)
//
Statements:
not necessarily \(S_1 \setminus (S_2 \setminus S_3) = (S_1 \setminus S_2) \setminus S_3\)
\(\land\)
\((S_1 \setminus S_2) \setminus S_3 \subseteq S_1 \setminus (S_2 \setminus S_3)\)
//
2: Natural Language Description
For any sets, \(S_1, S_2, S_3\), \(S_1 \setminus (S_2 \setminus S_3)\) is not necessarily \((S_1 \setminus S_2) \setminus S_3\), but \((S_1 \setminus S_2) \setminus S_3 \subseteq S_1 \setminus (S_2 \setminus S_3)\).
3: Proof
Whole Strategy: Step 1: see an example that \(S_1 \setminus (S_2 \setminus S_3) \neq (S_1 \setminus S_2) \setminus S_3\); Step 2: see that \((S_1 \setminus S_2) \setminus S_3 \subseteq S_1 \setminus (S_2 \setminus S_3)\).
Step 1:
For the 1st part, a counterexample suffices.
Let \(S_1 = \{0, 1\}, S_2 = \emptyset, S_3 = \{0\}\). Then, \(S_1 \setminus (S_2 \setminus S_3) = \{0, 1\} \setminus \emptyset = \{0, 1\}\); \((S_1 \setminus S_2) \setminus S_3 = \{0, 1\} \setminus \{0\} = \{1\}\).
Step 2:
For the 2nd part, for any \(p \in (S_1 \setminus S_2) \setminus S_3\), \(p \in S_1\), \(p \notin S_2\), so, \(p \notin S_2 \setminus S_3\), so, \(p \in S_1 \setminus (S_2 \setminus S_3)\).
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 set minus (set minus set) is union of 1st set minus 2nd set and intersection of 1st set and 3rd 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 any set minus (any set minus any set) is the union of the 1st set minus the 2nd set and the intersection of the 1st set and the 3rd 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(S_3\): \(\in \{\text{ the sets }\}\)
//
Statements:
\(S_1 \setminus (S_2 \setminus S_3) = (S_1 \setminus S_2) \cup (S_1 \cap S_3)\)
//
2: Natural Language Description
For any sets, \(S_1, S_2, S_3\), \(S_1 \setminus (S_2 \setminus S_3) = (S_1 \setminus S_2) \cup (S_1 \cap S_3)\).
3: Proof
Whole Strategy: Step 1: see that \(S_1 \setminus (S_2 \setminus S_3) \subseteq (S_1 \setminus S_2) \cup (S_1 \cap S_3)\); Step 2: see that \((S_1 \setminus S_2) \cup (S_1 \cap S_3) \subseteq S_1 \setminus (S_2 \setminus S_3)\).
Step 1:
For any \(p \in S_1 \setminus (S_2 \setminus S_3)\), \(p \in S_1\), \(p \notin S_2 \setminus S_3\), \(p \notin S_2\) or \(p \in S_3\), so, \(p \in S_1 \setminus S_2\) or \(p \in S_1 \cap S_3\), so, \(p \in (S_1 \setminus S_2) \cup (S_1 \cap S_3)\).
Step 2:
For any \(p \in (S_1 \setminus S_2) \cup (S_1 \cap S_3)\), \(p \in S_1\), \(p \notin S_2\) or \(p \in S_3\), so, \(p \notin S_2 \setminus S_3\), so, \(p \in S_1 \setminus (S_2 \setminus S_3)\).
4: Note
\(S_1 \setminus (S_2 \setminus S_3) = (S_1 \setminus S_2) \cup S_3\) does not necessarily hold, because for example, if \(S_3\) contains a point not contained in \(S_1\), the left hand does not contain the point but the right hand side contains the point.
\(S_1 \setminus (S_2 \setminus S_3)\) is not necessarily equal to but contains \((S_1 \setminus S_2) \setminus S_3\), as is proved in another article.
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 union of set minus set and set is not necessarily but contains union of 1st set and 3rd set minus union of 2nd set and 3rd 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 the union of any set minus any set and any set is not necessarily but contains the union of the 1st set and the 3rd set minus the union of the 2nd set and the 3rd 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(S_3\): \(\in \{\text{ the sets }\}\)
//
Statements:
not necessarily \((S_1 \setminus S_2) \cup S_3 = (S_1 \cup S_3) \setminus (S_2 \cup S_3)\)
\(\land\)
\((S_1 \cup S_3) \setminus (S_2 \cup S_3) \subseteq (S_1 \setminus S_2) \cup S_3\)
//
2: Natural Language Description
For any sets, \(S_1, S_2, S_3\), \((S_1 \setminus S_2) \cup S_3\) is not necessarily \((S_1 \cup S_3) \setminus (S_2 \cup S_3)\), but \((S_1 \cup S_3) \setminus (S_2 \cup S_3) \subseteq (S_1 \setminus S_2) \cup S_3\).
3: Proof
Whole Strategy: Step 1: see an example that \((S_1 \setminus S_2) \cup S_3 \neq (S_1 \cup S_3) \setminus (S_2 \cup S_3)\); Step 2: see that \((S_1 \cup S_3) \setminus (S_2 \cup S_3) \subseteq (S_1 \setminus S_2) \cup S_3\).
Step 1:
For the 1st part, a counterexample suffices.
Let \(S_1 = \emptyset\), \(S_2 = \emptyset\), \(S_3 \neq \emptyset\). \((S_1 \setminus S_2) \cup S_3 = S_3 \neq \emptyset\), but \((S_1 \cup S_3) \setminus (S_2 \cup S_3) = S_3 \setminus S_3 = \emptyset\).
Step 2:
For any \(p \in (S_1 \cup S_3) \setminus (S_2 \cup S_3)\), \(p \in S_1\) or \(p \in S_3\), but \(p \notin S_3\), so, \(p \in S_1\), \(p \notin S_2\), so, \(p \in S_1 \setminus S_2\), so, \(p \in (S_1 \setminus S_2) \cup S_3\).
4: Note
\((S_1 \setminus S_2) \cap S_3 = (S_1 \cap S_3) \setminus (S_2 \cap S_3)\) holds, as is proved in another article.
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 intersection of set minus set and set is intersection of 1st set and 3rd set minus intersection of 2nd set and 3rd 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 the intersection of any set minus any set and any set is the intersection of the 1st set and the 3rd set minus the intersection of the 2nd set and the 3rd 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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(S_2\): \(\in \{\text{ the sets }\}\)
\(S_3\): \(\in \{\text{ the sets }\}\)
//
Statements:
\((S_1 \setminus S_2) \cap S_3 = (S_1 \cap S_3) \setminus (S_2 \cap S_3)\)
//
2: Natural Language Description
For any sets, \(S_1, S_2, S_3\), \((S_1 \setminus S_2) \cap S_3 = (S_1 \cap S_3) \setminus (S_2 \cap S_3)\).
3: Proof
Whole Strategy: Step 1: see that \((S_1 \setminus S_2) \cap S_3 \subseteq (S_1 \cap S_3) \setminus (S_2 \cap S_3)\); Step 2: see that \((S_1 \cap S_3) \setminus (S_2 \cap S_3) \subseteq (S_1 \setminus S_2) \cap S_3\).
Step 1:
For any \(p \in (S_1 \setminus S_2) \cap S_3\), \(p \in S_1\) and \(p \in S_3\), so, \(p \in S_1 \cap S_3\), \(p \notin S_2\), \(p \notin S_2 \cap S_3\), so, \(p \in (S_1 \cap S_3) \setminus (S_2 \cap S_3)\).
Step 2:
For any \(p \in (S_1 \cap S_3) \setminus (S_2 \cap S_3)\), \(p \in S_1\), \(p \notin S_2 \cap S_3\), as \(p \in S_3\), \(p \notin S_2\), \(p \in S_1 \setminus S_2\), so, \(p \in (S_1 \setminus S_2) \cap S_3\).
4: Note
\((S_1 \setminus S_2) \cup S_3 = (S_1 \cup S_3) \setminus (S_2 \cup S_3)\) does not necessarily hold although \((S_1 \cup S_3) \setminus (S_2 \cup S_3) \subseteq (S_1 \setminus S_2) \cup S_3\) holds, as is proved in another article.
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 measurable map from measurable space into topological space
Topics
About:
measure
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of measurable map from measurable space into 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:
\( (S, A)\): \(\in \{\text{ the measurable spaces }\}\)
\( T\): \(\in \{\text{ the topological spaces }\}\) with topology \(O\)
\(*f\): \(: S \to T\)
//
Conditions:
\(\forall o \in O (f^{-1} (o) \in A)\)
//
2: Note
In fact, this concept is equivalent with 'measurable map from measurable space into topological space turned measurable space with the Borel \(\sigma\)-algebra': make \(T\) the measurable space, \((T, \sigma (O))\), where \(\sigma (O)\) is the Borel \(\sigma\)-algebra, then, any map, \(f: S \to T\), is measurable in this concept if and only if \(f\) is measurable from \((S, A)\) into \((T, \sigma (O))\).
That is because if \(f\) is measurable from \((S, A)\) into \((T, \sigma (O))\), \(f\) is measurable in this concept because \(O \subseteq \sigma (O)\), and if \(f\) is measurable in this concept, \(f\) is measurable from \((S, A)\) into \((T, \sigma (O))\) by the proposition that for any map between any measurable spaces, if the preimage of each element of any generator of the codomain \(\sigma\)-algebra is measurable, the map is measurable.
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 map between measurable spaces, if preimage of each element of generator of codomain \(\sigma\)-algebra is measurable, map is measurable
Topics
About:
measure
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 map between any measurable spaces, if the preimage of each element of any generator of the codomain \(\sigma\)-algebra is measurable, the map is measurable.
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:
\((S_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\((S_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(f\): \(: S_1 \to S_2\)
//
Statements:
(
\(\exists S \subseteq Pow (S_2) \text{ such that } A_2 = \sigma (S)\)
\(\land\)
\(\forall s \in S (f^{-1} (s) \in A_1)\)
)
\(\implies\)
\(f \in \{\text{ the measurable maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take the \(\sigma\)-algebra induced on \(S_2\) by \(f\), \(A\); Step 2: see that \(A_2 \subseteq A\); Step 3: see that the preimage of each element of \(A_2\) is measurable.
Step 1:
Let us take the \(\sigma\)-algebra induced on \(S_2\) by \(f\), \(A\).
Step 2:
By the supposition, \(S \subseteq A\).
As \(A_2 = \sigma (S)\) is the intersection of all the \(\sigma\)-algebras that contain \(S\), \(A_2 \subseteq A\), because \(A\) is a constituent of the intersection.
Step 3:
That means that for each \(a \in A_2\), \(f^{-1} (a) \in A_1\), which means that \(f\) is measurable.
References
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definition of \(\sigma\)-algebra induced on codomain of map from measurable space
Topics
About:
measure
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of \(\sigma\)-algebra induced on codomain of map from measurable 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:
\( (S_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\( S_2\): \(\in \{\text{ the sets }\}\)
\( f\): \(: S_1 \to S_2\)
\(*A_2\): \(= \{s \subseteq S_2: f^{-1} (s) \in A_1\}\), \(\in \{\text{ the } \sigma \text{ -algebras of } S_2\}\)
//
Conditions:
//
2: Note
Let us see that \(A_2\) is indeed a \(\sigma\)-algebra of \(S_2\).
1) \(S_2 \in A_2\): \(f^{-1} (S_2) = S_1 \in A_1\).
2) \(\forall a \in A_2 (S_2 \setminus a \in A_2)\): \(f^{-1} (S_2 \setminus a) = f^{-1} (S_2) \setminus f^{-1} (a)\), by the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset, \(= S_1 \setminus f^{-1} (a)\), but \(f^{-1} (a) \in A_1\), so, \(S_1 \setminus f^{-1} (a) \in A_1\).
3) \(\forall s: \mathbb{N} \to A_2 (\cup_{j \in \mathbb{N}} s (j) \in A_2)\): \(f^{-1} (\cup_{j \in \mathbb{N}} s (j)) = \cup_{j \in \mathbb{N}} f^{-1} (s (j))\), 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, but \(f^{-1} (s (j)) \in A_2\), so, \(\cup_{j \in \mathbb{N}} f^{-1} (s (j)) \in A_1\).
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 measurable map between measurable spaces
Topics
About:
measure
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of measurable map between measurable spaces.
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:
\( (S_1, A_1)\): \(\in \{\text{ the measurable spaces }\}\)
\( (S_2, A_2)\): \(\in \{\text{ the measurable spaces }\}\)
\(*f\): \(: S_1 \to S_2\)
//
Conditions:
\(\forall a_2 \in A_2 (f^{-1} (a_2) \in A_1)\)
//
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 Borel \(\sigma\)-algebra of topological space
Topics
About:
topological space
About:
measure
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of Borel \(\sigma\)-algebra of 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 topology \(O\)
\(*\sigma (O)\): \(= \text{ the } \sigma \text{ -algebra of } T \text{ generated by } O\)
//
Conditions:
//
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 \(\sigma\)-algebra of set generated by set of subsets
Topics
About:
measure
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of \(\sigma\)-algebra of set generated by set of subsets.
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:
\( S'\): \(\in \{\text{ the sets }\}\)
\( S\): \(\subseteq Pow S'\)
\(*\sigma (S)\): \(\in \{\text{ the } \sigma \text{ -algebras of } S'\}\)
//
Conditions:
\(S \subseteq \sigma (S)\)
\(\land\)
\(\forall A' \in \{\text{ the } \sigma \text{ -algebras of } S'\} \text{ such that } S \subseteq A' (\sigma (S) \subseteq A')\)
//
2: Note
In other words, \(\sigma (S)\) is the smallest \(\sigma\)-algebra that contains \(S\).
\(\sigma (S)\) is uniquely determined, because it is the intersection of all the \(\sigma\)-algebras of \(S'\) that contain \(S\), while at least, \(Pow S'\) is such a one: the intersection is a \(\sigma\)-algebra of \(S'\) that contains \(S\), by the proposition that for any set, the intersection of any \(\sigma\)-algebras is a \(\sigma\)-algebra; for each \(A' \in \{\text{ the } \sigma \text{ -algebras of } S'\} \text{ such that } S \subseteq A'\), \(\sigma (S) \subseteq A'\), because \(A'\) is a constituent of the intersection.
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 set, intersection of \(\sigma\)-algebras is \(\sigma\)-algebra
Topics
About:
measure
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, the intersection of any \(\sigma\)-algebras is a \(\sigma\)-algebra.
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:
\(S\): \(\in \{\text{ the sets }\}\)
\(\{A_j \vert j \in J\}\): \(J \in \{\text{ the possibly uncountable index sets }\}\), \(A_j \in \{\text{ the } \sigma \text{ -algebras of } S\}\)
\(A\): \(= \cap_{j \in J} A_j\)
//
Statements:
\(A \in \{\text{ the } \sigma \text{ -algebras of } S\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(A\) satisfies the requirements to be a \(\sigma\)-algebra.
Step 1:
1) \(S \in A\): for each \(j \in J\), \(S \in A_j\), so, \(S \in \cap_{j \in J} A_j = A\).
2) \(\forall a \in A (S \setminus a \in A)\): \(a \in A_j\) for each \(j \in J\), so, \(S \setminus a \in A_j\) for each \(j\), so, \(S \setminus a \in \cap_{j \in J} A_j = A\).
3) \(\forall s: \mathbb{N} \to A (\cup_{j \in \mathbb{N}} s (j) \in A)\): as \(s (k) \in A\), \(s (k) \in A_j\) for each \(j \in J\), so, \(s\) is \(: \mathbb{N} \to A_j\) for each \(j\), so, \(\cup_{k \in \mathbb{N}} s (k) \in A_j\) for each \(j\), so, \(\cup_{k \in \mathbb{N}} s (k) \in \cap_{j \in J} A_j = A\).
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 measurable space
Topics
About:
measure
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of measurable 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:
\(*(S, A)\): \(S \in \{\text{ the sets }\}\), \(A \in \{\text{ the } \sigma \text{ -algebras of } S\}\)
//
Conditions:
//
References
<The previous article in this series | The table of contents of this series | The next article in this series>
