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
- The reader knows a definition of 2nd-countable topological space.
- The reader knows a definition of completely regular topological space.
- The reader knows a definition of metrizable topological space.
- The reader knows a definition of continuous embedding.
- The reader admits 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 reader admits 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.
- The reader admits the proposition that for any disjoint subset and open set, the closure of the subset and the open set are disjoint.
- The reader admits the proposition that for any sequence of metric spaces with the induced topologies, this metric for the product set induces the product topology.
- The reader admits 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.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
- The reader admits the local criterion for openness.
- The reader admits the proposition that any closed continuous bijection is a homeomorphism.
- The reader admits 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.
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:
\(T\): \(\in \{\text{ the 2nd-countable topological spaces }\} \cap \{\text{ the completely regular topological spaces }\}\)
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Statements:
\(T \in \{\text{ the metrizable topological spaces }\}\)
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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.
