Borel Sigma-Algebra of Subset is Trace Sigma-Algebra

Theorem
Let $\left({X, \tau}\right)$ be a topological space, and let $A \in \tau$ be an open in $X$.

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.