1422: For Topological Space and Subspace, Borel \(\sigma\)-Algebra of Subspace Is Subspace \(\sigma\)-Algebra for Subspace of Space
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of Borel \(\sigma\)-algebra of topological space.
- The reader knows a definition of subspace \(\sigma\)-algebra of subset of measurable space.
- The reader admits the proposition that for any topological space, the inclusion from any subspace into the topological space is continuous.
- The reader admits the proposition that any continuous map between any topological spaces with the Borel \(\sigma\)-algebras is measurable.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any subspace, the Borel \(\sigma\)-algebra of the subspace is the subspace \(\sigma\)-algebra for the subspace of the 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:
\((T', O')\): \(\in \{\text{ the topological spaces }\}\) with the Borel \(\sigma\)-algebra, \(\sigma (O')\)
\((T, O)\): \(\subseteq T'\) with the subspace topology
\(\sigma (O)\): \(= \text{ the Borel } \sigma \text{ -algebra }\)
\(\sigma (O')_T\): \(= \text{ the subspace } \sigma \text{ -algebra for } T \text{ of } T'\)
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Statements:
\(\sigma (O) = \sigma (O')_T\)
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2: Proof
Whole Strategy: Step 1: think of the inclusion, \(\iota: T \to T'\), and see that \(\iota\) is measurable with respect to \(\sigma (O)\) and \(\sigma (O')\); Step 2: see that \(\sigma (O')_T \subseteq \sigma (O)\); Step 3: see that \(\sigma (O) \subseteq \sigma (O')_T\); Step 4: conclude the proposition.
Step 1:
Let us think of the inclusion, \(\iota: T \to T'\).
\(\iota\) is continuous, by the proposition that for any topological space, the inclusion from any subspace into the topological space is continuous.
\(\iota\) is measurable with respect to \(\sigma (O)\) and \(\sigma (O')\), by the proposition that any continuous map between any topological spaces with the Borel \(\sigma\)-algebras is measurable.
Step 2:
Let us see that \(\sigma (O')_T \subseteq \sigma (O)\).
Let \(a \in \sigma (O')_T\) be any.
\(a = a' \cap T\) where \(a' \in \sigma (O')\).
But \(a = a' \cap T = \iota^{-1} (a') \in \sigma (O)\), by Step 1.
Step 3:
\(\sigma (O')_T\) is a \(\sigma\)-algebra: refer to Note for the definition of subspace \(\sigma\)-algebra of subset of measurable space.
\(O \subseteq \sigma (O')_T\), because for each \(U \in O\), \(U = U' \cap T\) where \(U' \in O'\), but \(U' \in \sigma (O')\).
So, as \(\sigma (O)\) is the intersection of all the \(\sigma\)-algebras that contains \(O\), \(\sigma (O) \subseteq \sigma (O')_T\).
Step 4:
So, \(\sigma (O) = \sigma (O')_T\).
