Borel Sigma-Algebra of Subset is Trace Sigma-Algebra

Theorem
Let $\struct {X, \tau}$ 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:


 * $\map \BB {A, \tau_A} = \map \BB {X, \tau}_A$

where $\BB$ signifies Borel $\sigma$-algebra, and $\map \BB {X, \tau}_A$ signifies trace $\sigma$-algebra.

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


 * $\map \BB {X, \tau} = \map \sigma \tau$

and also, by definition of subspace topology:


 * $\tau_A = A \cap \tau = \set {A \cap U: U \in \tau}$

Thus, it follows that:


 * $\map \BB {A, \tau_A} = \map \sigma {A \cap \tau}$

Thereby, the desired equality:


 * $\map \BB {A, \tau_A} = \map \BB {X, \tau}_A$

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