Borel Sigma-Algebra of Subset is Trace Sigma-Algebra

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Theorem

Let $\left({X, \tau}\right)$ be a topological space, and let $A \subseteq X$ be a subset of $X$.

Let $\tau_A$ be the subspace topology on $A$.


Then the following equality of $\sigma$-algebras on $A$ holds:

$\mathcal B \left({A, \tau_A}\right) = \mathcal B \left({X, \tau}\right)_A$

where $\mathcal B$ signifies Borel $\sigma$-algebra, and $\mathcal B \left({X, \tau}\right)_A$ signifies trace $\sigma$-algebra.


Proof

By definition of Borel $\sigma$-algebra, it holds that:

$\mathcal B \left({X, \tau}\right) = \sigma \left({\tau}\right)$

and also, by definition of subspace topology:

$\tau_A = A \cap \tau = \left\{{A \cap U: U \in \tau}\right\}$

Thus, it follows that:

$\mathcal B \left({A, \tau_A}\right) = \sigma \left({A \cap \tau}\right)$


Thereby, the desired equality:

$\mathcal B \left({A, \tau_A}\right) = \mathcal B \left({X, \tau}\right)_A$

follows directly from applying Trace Sigma-Algebra of Generated Sigma-Algebra with $\mathcal G = \tau$.

$\blacksquare$


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