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description/proof of that finite product of topological sums is topological sum of products
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 product of any topological sums is the topological sum of the products of the constituents of the topological sums.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{L_j \in \{\text{ the index sets }\} \vert j \in J\}\):
\(\{T_{j, l_j} \in \{\text{ the topological spaces }\} \vert j \in J, l_j \in L_j\}\)
\(T\): \(= \times_{j \in J} \coprod_{l_j \in L_j} T_{j, l_j}\)
\(T'\): \(= \coprod_{\times_{j \in J} l_j \in \times_{j \in J} L_j} \times_{j \in J} T_{j, l_j}\)
//
Statements:
\(T = T'\)
//
2: Proof
Whole Strategy: Step 1: see that \(T\) and \(T'\) are the same set-wise; Step 2: see that each open \(U \subseteq T\) is open on \(T'\); Step 3: see that each open \(U' \subseteq T'\) is open on \(T\).
Step 1:
Let \(J = \{1, ..., n\}\) without loss of generality, just for our convenience of expressions.
\(T\) and \(T'\) are the same set-wise, because there is the identity map, \(id: T \to T', (p_1, ..., p_n) \mapsto (p_1, ..., p_n)\), where \((p_1, ..., p_n) \in T\) means that \(p_1 \in T_{1, l_1}, ..., p_n \in T_{n, l_n}\), so, \((p_1, ..., p_n) \in \times_{j \in J} T_{j, l_j}\) in \(T'\), while it is indeed an injection, because for any \((p_1, ..., p_n) \neq (p'_1, ..., p'_n)\), \(p_j \neq p'_j\) for a \(j \in J\), and \(id ((p_1, ..., p_n))^j = p_j \neq p'_j = id ((p'_1, ..., p'_n))^j\), and is a surjection, because for each \((p_1, ..., p_n) \in T'\), \(id ((p_1, ..., p_n)) = (p_1, ..., p_n)\).
The issue is whether \(T\) and \(T'\) have the same topology.
Step 2:
Let \(U \subseteq T\) be any open subset.
\(U = \cup_{m \in M} \times_{j \in j} U_{j, m}\), where \(M\) is a possibly uncountable index set and \(U_{j, m} \subseteq \coprod_{l_j \in L_j} T_{j, l_j}\) is open on \(\coprod_{l_j \in L_j} T_{j, l_j}\), by Note for the definition of product topology.
\(U \cap \times_{j \in J} T_{j, l_j} = (\cup_{m \in M} \times_{j \in j} U_{j, m}) \cap (\times_{j \in J} T_{j, l_j}) = \cup_{m \in M} (\times_{j \in j} U_{j, m} \cap \times_{j \in J} T_{j, l_j})\), 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, \(= \cup_{m \in M} \times_{j \in J} (U_{j, m} \cap T_{j, l_j})\), by the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets.
For each \(j \in J\), \(U_{j, m} \cap T_{j, l_j}\) is open on \(T_{j, l_j}\), because \(U_{j, m}\) is open on \(\coprod_{l_j \in L_j} T_{j, l_j}\), by the definition of topological sum.
So, \(\times_{j \in J} (U_{j, m} \cap T_{j, l_j})\) is open on \(\times_{j \in J} T_{j, l_j}\), by the definition of product topology, so, \(U \cap \times_{j \in J} T_{j, l_j}\) is open on \(\times_{j \in J} T_{j, l_j}\).
So, \(U\) is open on \(T'\), by the definition of topological sum.
Step 3:
Let \(U' \subseteq T'\) be any open subset.
\(U' \cap (\times_{j \in J} T_{j, l_j})\) is open on \(\times_{j \in J} T_{j, l_j}\), by the definition of topological sum.
So, \(U' \cap (\times_{j \in J} T_{j, l_j}) = \cup_{m \in M_{\times_{j \in J} l_j}} \times_{j \in J} U_{j, m}\), where \(M_{\times_{j \in J} l_j}\) is a possibly uncountable index set and \(U_{j, m} \subseteq T_{j, l_j}\) is open on \(T_{j, l_j}\), by Note for the definition of product topology.
\(U' = \cup_{\times_{j \in J} l_j \in \times_{j \in J} L_j} (U' \cap (\times_{j \in J} T_{j, l_j})) = \cup_{\times_{j \in J} l_j \in \times_{j \in J} L_j} \cup_{m \in M_{\times_{j \in J} l_j}} \times_{j \in J} U_{j, m}\).
But \(U_{j, m}\) is open on \(\coprod_{l_j \in L_j} T_{j, l_j}\), because \(U_{j, m} \cap T_{j, l_j} = U_{j, m}\) is open on \(T_{j, l_j}\) and \(U_{j, m} \cap T_{j, l'_j} = \emptyset\) for any \(l'_j \neq l_j\), by the definition of topological sum.
So, \(\times_{j \in J} U_{j, m}\) is open on \(T\), by the definition of product topology, and \(U' = \cup_{\times_{j \in J} l_j \in \times_{j \in J} L_j} \cup_{m \in M_{\times_{j \in J} l_j}} \times_{j \in J} U_{j, m}\) is open on \(T\).
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 injective \(C^\infty\) immersion is C^\infty embedding iff restriction of map on range codomain is open
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 any injective \(C^\infty\) immersion is a \(C^\infty\) embedding if and only if the restriction of the map on the range codomain is open.
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:
\(M_1\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(M_2\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the injective } C^\infty \text{ immersions }\}\)
\(f'\): \(: M_1 \to f (M_1), m_1 \mapsto f (m_1)\), where \(f (M_1) \subseteq M_2\) has the subspace topology
//
Statements:
\(f \in \{\text{ the } C^\infty \text{ embeddings }\}\)
\(\iff\)
\(f' \in \{\text{ the open maps }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f\) is locally a \(C^\infty\) embedding, and for each \(m_1 \in M_1\), take an open neighborhood of \(m_1\), \(U_{m_1}\), such that \(f'_{m_1} := f \vert_{U_{m_1}}: U_{m_1} \to f (U_{m_1})\) is a homeomorphism; Step 2: suppose that \(f'\) is open; Step 3: see that \(f'^{-1}\) is continuous; Step 4; suppose that \(f\) is a \(C^\infty\) embedding; Step 5: see that \(f'\) is open.
Step 1:
As \(f\) is injective, \(f'\) is a bijection, so, there is the inverse, \(f'^{-1}: f (M_1) \to M_1\).
As \(f\) is a \(C^\infty\) immersion, \(f\) is locally an \(C^\infty\) embedding, by the proposition that any \(C^\infty\) immersion is locally a \(C^\infty\) embedding, which means that for any point, \(m_1 \in M_1\), there is an open neighborhood of \(m_1\), \(U_{m_1}\), such that \(f'_{m_1} := f \vert_{U_{m_1}}: U_{m_1} \to f (U_{m_1})\) is a homeomorphism with the inverse, \({f'_{m_1}}^{-1}: f (U_{m_1}) \to U_{m_1}\): while \(f (U_{m_1})\) is a topological subspace of \(M_2\), it is a topological subspace of \(f (M_1)\), by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Also \({f'_{m_1}}^{-1}: f (U_{m_1}) \to M_1\) is continuous, by the proposition that any expansion of any continuous map on the codomain is continuous, which is a domain restriction of \(f'^{-1}\), while \(\{f (U_{m_1})\}\) is a (not necessarily open) cover of \(f (M_1)\).
Step 2:
Let us suppose that \(f'\) is open.
Step 3:
\(\{f (U_{m_1})\}\) is an open cover of \(f (M_1)\).
As the domain restriction of \(f'^{-1}\) to each element of the open cover is continuous, \(f'^{-1}\) 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, so, \(f'\) is a homeomorphism, so, \(f\) is a \(C^\infty\) embedding.
Step 4:
Let us suppose that \(f\) is a \(C^\infty\) embedding.
Step 5:
\(f'\) is a homeomorphism. \(f'^{-1}\) is continuous.
For any open subset, \(U \subseteq M_1\), \({f'^{-1}}^{-1} (U) = f' (U)\) is open on \(f (M_1)\), so, \(f'\) is open.
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 metric space with induced topology, compact subset, and open cover of subset, there is positive real number (Lebesgue number) s.t. subset of subset with diameter smaller than number is contained in element of cover
Topics
About:
metric space
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 metric space with the induced topology, any compact subset, and any open cover of the subset, there is a positive real number (Lebesgue number) such that each subset of the subset with diameter smaller than the number is contained in an element of the cover.
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:
\(M\): \(\in \{\text{ the metric spaces }\}\), with the induced topology
\(S\): \(\in \{\text{ the compact subsets of } M\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(Q\): \(= \{U_j \in \{\text{ the open subsets of } M\} \vert j \in J\}\) such that \(S \subseteq \cup_{j \in J} U_j\)
//
Statements:
\(\exists r \in \mathbb{R} \text{ such that } 0 \lt r (\forall S^` \subseteq S \text{ such that } Diam (S^`) \lt r (\exists j \in J (S^` \subseteq U_j)))\)
//
2: Note
When \(M\) is compact, \(S = M\) is a compact subset, and for each \(Q\), there is an \(r\) such that for each \(S^` \subseteq M\) with diameter smaller than \(r\), there is a \(j \in J\) such that \(S^` \subseteq U_j\).
3: Proof
Whole Strategy: Step 1: take an open cover of \(S\), \(\{B_{s, \epsilon_s / 2} \vert s \in S\}\), such that \(B_{s, \epsilon_s} \subseteq U_j\), and take a finite subcover, \(\{B_{s_l, \epsilon_{s_l} / 2} \vert l \in L\}\); Step 2: take \(r := Min (\{\epsilon_{s_l} / 2 \vert l \in L\})\); Step 3: see that \(r\) satisfies the conditions for this proposition.
Step 1:
For each \(s \in S\), \(s \in U_j\) for a \(j \in J\), and there is a \(B_{s, \epsilon_s} \subseteq M\) such that \(B_{s, \epsilon_s} \subseteq U_j\), where \(\epsilon_s\) depends on \(s\).
\(\{B_{s, \epsilon_s / 2} \vert s \in S\}\) is an open cover of \(S\).
As \(S\) is compact, there is a finite subcover, \(\{B_{s_l, \epsilon_{s_l} / 2} \vert l \in L\}\).
Step 2:
Let us take \(r := Min (\{\epsilon_{s_l} / 2 \vert l \in L\})\).
Step 3:
Let us see that \(r\) satisfies the conditions for this proposition.
Let \(S^`\) be any.
When \(S^` = \emptyset\), for any \(j \in J\), \(S^` \subseteq U_j\).
Let us suppose otherwise, hereafter.
There is an \(s^` \in S^`\).
\(s^` \in B_{s_l, \epsilon_{s_l} / 2}\) for an \(l \in L\), but \(B_{s_l, \epsilon_{s_l} / 2} \subseteq B_{s_l, \epsilon_{s_l}} \subseteq U_j\) for a \(j \in J\).
For each \({s^`}' \in S^`\), \(dist (s_l, {s^`}') \le dist (s_l, s^`) + dist (s^`, {s^`}') \lt \epsilon_{s_l} / 2 + r \le \epsilon_{s_l} / 2 + \epsilon_{s_l} / 2 = \epsilon_{s_l}\), so, \({s^`}' \in B_{s_l, \epsilon_{s_l}} \subseteq U_j\).
So, \(S^` \subseteq U_j\).
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 subspace of completely regular topological space is completely regular
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 subspace of any completely regular topological space is completely regular.
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:
\(T'\): \(\in \{\text{ the completely regular topological spaces }\}\)
\(T\): \(\subseteq T'\) with the subspace topology
//
Statements:
\(T \in \{\text{ the completely regular topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: for each \(t \in T\) and each \(C \subseteq T\) such that \(t \notin C\), take \(C' \subseteq T'\) such that \(C = C' \cap T\), take \(f': T' \to [0, 1]\) such that \(f' (t) = 0\) and \(f' (C') = \{1\}\), and see that \(f := f' \vert_T\) will do.
Step 1:
Let \(t \in T\) be any.
Let \(C \subseteq T\) be any closed subset such that \(t \notin C\).
\(C = C' \cap T\) for a closed \(C' \subseteq T'\), by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
While \(t \in T'\), \(t \notin C'\), because if \(t \in C'\), \(t \in C' \cap T = C\), a contradiction.
As \(T'\) is completely regular, there is a continuous \(f': T' \to [0, 1]\) such that \(f' (t) = 0\) and \(f' (C') = \{1\}\).
Let us take \(f := f' \vert_T = T \to [0, 1]\).
\(f\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
\(f (t) = 0\).
\(f (C) = \{1\}\), because \(C \subseteq C'\) and \(f (C) = f' (C)\).
So, \(T\) is completely regular.
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 finite-product topological space, product of constituent subbases is subbasis
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 finite-product topological space, the product of any constituent subbases is a subbasis.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{T_j \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_j\): \(= \text{ the product topological space }\)
\(\{S_j \in \{\text{ the subbases for } T_j\} \vert j \in J\}\):
\(S\): \(= \{\times_{j \in J} s_j \vert s_j \in S_j\}\)
//
Statements:
\(S \in \{\text{ the subbases for } \times_{j \in J} T_j\}\)
//
2: Note
It needs to be a finite-product, because this is based on the proposition that for any finite-product topological space, the product of any constituent bases is a basis.
3: Proof
Whole Strategy: Step 1: see that \(S\) satisfies the conditions to be a subbasis.
Step 1:
For each \(j \in J\), let \(B_j\) be the set of the intersections of the finite subsets of \(S_j\), which is a basis for \(T_j\), by the definition of subbasis for topological space.
\(B := \{\times_{j \in J} b_j \vert b_j \in B_j\}\) is a basis for \(\times_{j \in J} T_j\), by the proposition that for any finite-product topological space, the product of any constituent bases is a basis.
Let us think of the intersection of any finite subset of \(S\), \(\times_{j \in J} s_{1, j} \cap ... \cap \times_{j \in J} s_{n, j}\).
\(\times_{j \in J} s_{1, j} \cap ... \cap \times_{j \in J} s_{n, j} = \times_{j \in J} (s_{1, j} \cap ... \cap s_{n, j})\), by the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets.
For each \(j \in J\), \(s_{1, j} \cap ... \cap s_{n, j}\) is an element of \(B_j\).
So, \(\times_{j \in J} (s_{1, j} \cap ... \cap s_{n, j}) \in B\).
That means that the set of the intersections of the finite subsets of \(S\) is contained in \(B\).
Let \(J = \{1, ..., m\}\) without loss of generality, just for our convenience of expressions.
Each element of \(B\) is \((s_{1, 1} \cap ... \cap s_{n_1, 1}) \times ... \times (s_{1, m} \cap ... \cap s_{n_m, m})\).
Let us take \(n := Max (\{n_1, ..., n_m\})\).
For each \(j \in J\), if \(n_j \lt n\), let \(s_{n_j + 1, j} = ... = s_{n, j} = s_{n_j, j}\).
Then, \((s_{1, 1} \cap ... \cap s_{n_1, 1}) \times ... \times (s_{1, m} \cap ... \cap s_{n_m, m}) = (s_{1, 1} \cap ... \cap s_{n, 1}) \times ... \times (s_{1, m} \cap ... \cap s_{n, m})\).
\(= \times_{j \in J} s_{1, j} \cap ... \cap \times_{j \in J} s_{n, j}\), by the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets, which is an element of the set of the intersections of the finite subsets of \(S\).
That means that \(B\) is contained in the set of the intersections of the finite subsets of \(S\).
So, the set of the intersections of the finite subsets of \(S\) is nothing but \(B\), a basis.
So, \(S\) is a subbasis for \(\times_{j \in J} T_j\).
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 finite-product topological space, product of constituent bases is basis
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 finite-product topological space, the product of any constituent bases is a basis.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{T_j \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_j\): \(= \text{ the product topological space }\)
\(\{B_j \in \{\text{ the bases for } T_j\} \vert j \in J\}\):
\(B\): \(= \{\times_{j \in J} b_j \vert b_j \in B_j\}\)
//
Statements:
\(B \in \{\text{ the bases for } \times_{j \in J} T_j\}\)
//
2: Note
It needs to be a finite-product, because otherwise, \(\times_{j \in J} b_j\) would not be open in general, because not only some finite of \(b_j\) s would not be \(T_j\) s.
3: Proof
Whole Strategy: Step 1: see that \(B\) satisfies the conditions to be a basis.
Step 1:
Each \(\times_{j \in J} b_j \in B\) is open on \(\times_{j \in J} T_j\), because each \(b_j \subseteq T_j\) is open and only finite of \(b_j\) s are not \(T_j\) s (there are only finite of \(b_j\) s).
Let \(t \in \times_{j \in J} T_j\) be any.
Let \(N_t \subseteq \times_{j \in J} T_j\) be any neighborhood of \(t\).
There is a \(\times_{j \in J} U_j \subseteq \times_{j \in J} T_j\) such that \(t \in \times_{j \in J} U_j \subseteq N_t\), where \(U_j \subseteq T_j\) is an open subset, by the definition of product topology.
For each \(j \in J\), \(t^j \in U_j\), so, there is a \(b_j \in B_j\) such that \(t^j \in b_j \subseteq U_j\), by the definition of basis for topological space.
\(\times_{j \in J} b_j \in B\) and \(t \in \times_{j \in J} b_j \subseteq \times_{j \in J} U_j \subseteq N_t\).
So, \(B\) is a basis for \(\times_{j \in J} T_j\).
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 \(1\)-dimensional Euclidean topological space, set of upper bounded open intervals and lower bounded open intervals is subbasis
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, the set of the upper bounded open intervals and the lower bounded open intervals is a subbasis.
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:
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(S\): \(= \{(- \infty, r) \subseteq \mathbb{R} \vert r \in \mathbb{R}\} \cup \{(r, \infty) \subseteq \mathbb{R} \vert r \in \mathbb{R}\}\)
//
Statements:
\(S \in \{\text{ the subbases for } \mathbb{R}\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(S\) satisfies the conditions to be a subbasis.
Step 1:
The intersection of each finite subset of \(S\) is open, as a finite intersection of open subsets.
Let \(r \in \mathbb{R}\) be any.
Let \(N_r \subseteq \mathbb{R}\) be any neighborhood of \(r\).
There is a \(B_{r, \epsilon} \subseteq \mathbb{R}\) such that \(r \in B_{r, \epsilon} \subseteq N_r\).
\(B_{r, \epsilon} = (- \infty, r + \epsilon) \cap (r - \epsilon, \infty)\).
So, \(r \in (- \infty, r + \epsilon) \cap (r - \epsilon, \infty) \subseteq N_r\).
So, the set of the intersections of the finite subsets of \(S\) is a basis for \(\mathbb{R}\).
So, \(S\) is a subbasis for \(\mathbb{R}\).
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 \(2\) distinct non-negative real numbers and natural number larger than \(1\), there are 2nd and 3rd natural numbers s.t. 3rd number divided by number to power of 2nd number is exactly between real numbers
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 \(2\) distinct non-negative real numbers and any natural number larger than \(1\), there are some 2nd and 3rd natural numbers such that the 3rd number divided by the number to the power of the 2nd number is exactly between the real numbers.
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:
\(\{r_1, r_2\}\): \(\subseteq \mathbb{R}\), such that \(0 \le r_1, r_2 \land r_1 \lt r_2\)
\(n\): \(\in \mathbb{N} \setminus \{0, 1\}\)
//
Statements:
\(\exists m, j \in \mathbb{N} (r_1 \lt j / n^m \lt r_2)\)
//
2: Note
\(n \lt 1\) is required, because if \(n = 1\), \(n^m = 1\) for whatever \(m\) and \(j / n^m = j / 1 = j\) could not satisfy \(r_1 \lt j / n^m \lt r_2\) for \(r_1, r_2 \lt 1\).
3: Proof
Whole Strategy: Step 1: take an \(m\) such that \(1 \lt n^m (r_2 - r_1)\) and take a \(j\) such that \(n^m r_1 \lt j \lt n^m r_2\).
Step 1:
As \(0 \lt r_2 - r_1\) and \(n^m\) can be any larger by choosing a large \(m\) (because \(1 \lt n\)), let us choose an \(m \in \mathbb{N}\) such that \(1 \lt n^m (r_2 - r_1) = n^m r_2 - n^m r_1\).
There is a \(j \in \mathbb{N}\) such that \(n^m r_1 \lt j \lt n^m r_2\): when \(n^m r_1\) is a natural number, \(j := n^m r_1 + 1 \lt n^m r_2\) will do; otherwise, \(j\) can be taken to be the smallest natural number larger than \(n^m r_1\), then, \(j \lt n^m r_2\).
Then, \(r_1 \lt j / n^m \lt r_2\).
References
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description/proof of that 2nd-countable completely regular topological space is metrizable (Urysohn metrization theorem)
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 2nd-countable completely regular topological space is metrizable (Urysohn metrization theorem).
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:
\(T\): \(\in \{\text{ the 2nd-countable topological spaces }\} \cap \{\text{ the completely regular topological spaces }\}\)
//
Statements:
\(T \in \{\text{ the metrizable topological spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: take a countable set of some continuous maps into the closed unit interval such that for each point and each closed subset that does not contain the point, there is an element of the set that is \(0\) over an open neighborhood of the point and is \(1\) over the closed subset, \(\{f_j \vert j \in \mathbb{N} \setminus \{0\}\}\); Step 2: define \(f: T \to \times_{j \in \mathbb{N} \setminus \{0\}} [0, 1], t \mapsto \times_{j \in \mathbb{N} \setminus \{0\}} f_j (t)\), and see that \(f\) is a continuous embedding while \(f (T) \subseteq \times_{j \in \mathbb{N} \setminus \{0\}} [0, 1]\) as the topological subspace has the topology induced by the submetric of a metric on \(\times_{j \in \mathbb{N} \setminus \{0\}} [0, 1]\) that induces the product topology; Step 3: see that \(T\) is induced by the metric induced by the homeomorphism from the submetric.
Step 1:
Let us take a countable set of some continuous maps into the closed unit interval such that for each point and each closed subset that does not contain the point, there is an element of the set that is \(0\) over an open neighborhood of the point and is \(1\) over the closed subset, \(\{f_j \vert j \in \mathbb{N} \setminus \{0\}\}\), by the proposition that for any 2nd-countable completely regular topological space, there is a countable set of some continuous maps into the closed unit interval such that for each point and each closed subset that does not contain the point, there is an element of the set that is \(0\) over an open neighborhood of the point and is \(1\) over the closed subset: the countable index set in the proposition, \(J\), can be taken as \(\mathbb{N} \setminus \{0\}\), because when \(J\) is infinite countable, there is a bijection from \(\mathbb{N} \setminus \{0\}\) onto \(J\), and otherwise, \(\{f_j \vert j \in \{1, ..., n\}\}\) can be augmented such that \(f_j = f_n\) for each \(n \lt j\) (in fact, the finite case does not need to be made infinite, but for our convenience of making the descriptions uniform, it is done).
Step 2:
Let \([0, 1] \subseteq \mathbb{R}\) be the topological subspace of the Euclidean \(\mathbb{R}\).
\(\times_{j \in \mathbb{N} \setminus \{0\}} [0, 1]\) be the product topological space.
Let us define \(f: T \to \times_{j \in \mathbb{N} \setminus \{0\}} [0, 1], t \mapsto \times_{j \in \mathbb{N} \setminus \{0\}} f_j (t)\).
Let us see that \(f\) is a continuous embedding.
Let \(\pi^j: \times_{j \in \mathbb{N} \setminus \{0\}} [0, 1] \to [0, 1]\) be the projection onto the \(j\)-constituent.
For each \(j \in \mathbb{N} \setminus \{0\}\), \(\pi^j \circ f: T \to [0, 1]\) is continuous, because it is nothing but \(f_j\).
\(f\) is continuous, by the proposition that any map from any topological space into any product topological space is continuous if and only if each component map is continuous.
\(f\) is injective, because for each \(t, t' \in T\) such that \(t \neq t'\), there are an open neighborhood of \(t\), \(U_t \subseteq T\), and an open neighborhood of \(t'\), \(U_{t'} \subseteq T\), such that \(U_t \cap U_{t'} = \emptyset\), then, \(U_t \cap \overline{U_{t'}} = \emptyset\), by the proposition that for any disjoint subset and open set, the closure of the subset and the open set are disjoint, so, \(t \notin \overline{U_{t'}}\), so, there is a \(f_j\) such that \(f_j (t) = 0\) and \(f_j (\overline{U_{t'}}) = 1\), which implies that \(f_j (t) \neq f_j (t')\), so, \(f (t) \neq f (t')\).
So, the codomain restriction of \(f\), \(f': T \to f (T)\), is a bijection.
\(\times_{j \in \mathbb{N} \setminus \{0\}} [0, 1]\) has the topology that is induced by a metric, \(dist'\), by the proposition that for any sequence of metric spaces with the induced topologies, this metric for the product set induces the product topology.
\(f (T) \subseteq \times_{j \in \mathbb{N} \setminus \{0\}} [0, 1]\) as the topological subspace has the topology induced by the submetric of \(dist'\), by the proposition that for any topological space induced by any metric and any subset, the subset as the topological subspace equals the subset as the topological space induced by the metric subspace.
\(f'\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
Let us see that \(f'\) is closed.
Let \(C \subseteq T\) be any closed subset.
Let us see that \(f (T) \setminus f (C) \subseteq f (T)\) is open.
Let \(p \in f (T) \setminus f (C)\) be any.
\(p = f (t)\) for a \(t \in T \setminus C\), so, \(t \notin C\).
There is a \(f_l\) such that \(f_l (t) = 0\) and \(f_l (C) = \{1\}\), so, \(p^l = f (t)^l = f_l (t) = 0\).
There is an open neighborhood of \(p^l\), \(U'_{p^l} \subseteq [0, 1]\), such that \(U'_{p^l} \cap f_l (C) = \emptyset\).
Let \(U'_p := \times_{j \in J} U'_{p^j}\) where \(U'_{p^j} = [0, 1]\) for each \(j \in \mathbb{N} \setminus \{0, l\}\), which is an open neighborhood of \(p\) on \(\times_{j \in \mathbb{N} \setminus \{0\}} [0, 1]\).
Let \(U_p := U'_p \cap f (T)\), which is an open neighborhood of \(p\) on \(f (T)\).
\(U_p \subseteq f (T) \setminus f (C)\), because for each \(p' \in U_p\), \(p' \in f (T)\), \(p'^l \in U'_{p^l}\), so, \(p'^l \notin f_l (C)\), so, \(p' \notin f (C)\), so, \(p' \in f (T) \setminus f (C)\).
So, by the local criterion for openness, \(f (T) \setminus f (C) \subseteq f (T)\) is open, so, \(f (C) \subseteq f (T)\) is closed.
So, \(f'\) is closed.
So, \(f'\) is a homeomorphism, by the proposition that any closed continuous bijection is a homeomorphism.
Step 3:
As \(f (T)\) is induced by the submetric of \(dist'\), \(T\) is induced by the metric induced by \(f'\) from the submetric, by the proposition that for any homeomorphism from any metric space with the induced topology, the codomain topology is induced by the metric induced by the homeomorphism.
So, \(T\) is metrizable.
References
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definition of metrizable topological space
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of metrizable 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 the topology, \(O\)
//
Conditions:
\(\exists dist \in \{\text{ the metrics for } T\} (\text{ the topology induced by } dist = O)\)
//
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 2nd-countable completely regular topological space, there is countable set of continuous maps into closed unit interval s.t. for each point and each closed subset that does not contain point, there is element of set that is \(0\) over open neighborhood of point and is \(1\) over closed 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 2nd-countable completely regular topological space, there is a countable set of some continuous maps into the closed unit interval such that for each point and each closed subset that does not contain the point, there is an element of the set that is \(0\) over an open neighborhood of the point and is \(1\) over the closed subset.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the 2nd-countable topological spaces }\} \cap \{\text{ the completely regular topological spaces }\}\)
\([0, 1]\): \(\subseteq \mathbb{R}\), with the subspace topology
//
Statements:
\(\exists J \in \{\text{ the countable index sets }\}, \exists S = \{f_j: T \to [0, 1] \in \{\text{ the continuous maps }\} \vert j \in J\} (\forall t \in T (\forall C \in \{\text{ the closed subsets of } T\} \text{ such that } t \notin C (\exists f_j \in S (\exists U_t \in \{\text{ the open neighborhoods of } t\} (f (U_t) = \{0\} \land f (C) = \{1\})))))\)
//
2: Proof
Whole Strategy: Step 1: take any countable basis, \(B\), take the set of the pairs of some elements of \(B\), \(J := \{(b_1, b_2)\}\), such that there is a continuous map, \(f: T \to [0, 1]\), such that \(f (b_2) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\), choose such an \(f\) for each such pair, and take the set, \(S' := \{((b_1, b_2), f_{(b_1, b_2)}) \vert (b_1, b_2) \in J\}\); Step 2: take \(S := \{f_j \vert j \in J, (j, f_j) \in S'\}\); Step 3: see that \(S\) satisfies the requirements of this proposition.
Step 0:
When \(T = \emptyset\), the proposition vacuously holds, because \(J = \mathbb{N}\) and for each \(j \in J\), the only possible \(f_j: T \to [0, 1]\) will do, because there is no \(t \in T\).
Let us suppose otherwise, hereafter.
Step 1:
Let \(B\) be any countable basis for \(T\).
Let \(J \subseteq B \times B\) be the set of the pairs of some elements of \(B\) such that for each \((b_1, b_2) \in J\), there is a continuous map, \(f: T \to [0, 1]\), such that \(f (b_2) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\).
\(J\) is countable, because it is a subset of the countable, \(B \times B\).
\(J\) is nonempty, in fact, for each nonempty \(b_1 \in B\), there is at least \(1\) \(b_2 \in B\) such that \((b_1, b_2) \in J\), because there is a \(t \notin T \setminus b_1\), so, there is an open neighborhood of \(t\), \(U_t \subseteq T\), and a continuous map, \(f: T \to [0, 1]\), such that \(f (U_t) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\), because \(T\) is completely regular (refer to Note for the definition of completely regular topological space), and there is a \(b_2 \in B\) such that \(t \in b_2 \in U_t\), so, \(f (b_2) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\) hold.
Now, we have the function, \(F\), that the domain is \(J\) and for each \((b_1, b_2) \in J\), \(F ((b_1, b_2)) = \{f: T \to [0, 1] \in \{\text{ the continuous maps }\} \vert f (b_2) = \{0\} \land f (T \setminus b_1) = \{1\}\}\), which is nonempty.
By the axiom of choice, there is a function, \(F'\), that the domain is \(J\) and for each \((b_1, b_2) \in J\), \(F' ((b_1, b_2)) \in F ((b_1, b_2))\).
Let us take \(S' := \{((b_1, b_2), f_{(b_1, b_2)} := F' ((b_1, b_2))) \vert (b_1, b_2) \in J\}\).
Step 2:
Let us take \(S := \{f_j \vert j \in J, (j, f_j) \in S'\}\).
\(S\) is a countable set.
Step 3:
Let us see that \(S\) satisfies the requirements of this proposition.
Let \(t \in T\) be any.
Let \(C \subseteq T\) be any closed subset such that \(t \notin C\).
\(t \in T \setminus C\), so, \(T \setminus C \subseteq T\) is an open neighborhood of \(t\).
So, there is a \(b_1 \in B\) such that \(t \in b_1 \subseteq T \setminus C\).
\(t \notin T \setminus b_1\).
There are an open neighborhood of \(t\), \(U_t \subseteq T\), and a continuous \(f: T \to [0, 1]\) such that \(f (U_t) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\), because \(T\) is completely regular.
There is a \(b_2 \in B\) such that \(t \in b_2 \subseteq U_t\).
So, \(f (b_2) = \{0\}\) and \(f (T \setminus b_1) = \{1\}\).
So, \((b_1, b_2) \in J\).
So, \(f_{(b_1, b_2)} \in S\) satisfies that \(f_{(b_1, b_2)} (b_2) = \{0\}\) and \(f_{(b_1, b_2)} (T \setminus b_1) = \{1\}\).
\(C \subseteq T \setminus b_1\), because for each \(c \in C\), \(c \notin T \setminus C\), so, \(c \notin b_1 \subseteq T \setminus C\), so, \(c \in T \setminus b_1\), so, \(f_{(b_1, b_2)} (C) = \{1\}\).
References
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