Borel $\sigma$-algebra of subset of a topological space

Question

I've tried to prove the following for several hours now but struggle to finish it (maybe it is super obvious/easy, but I don't have the right thoughts):

Let $(\Omega, \mathcal T)$ be a topological space, $\Omega' \in \sigma(\mathcal T, \Omega) = \mathcal B(\Omega, \mathcal T)$ and $\mathcal T'$ the subspace topology on $\Omega'$ with respect to $\mathcal T$. Show that $$\mathcal B(\Omega', \mathcal T') = \{A\in \mathcal B(\Omega, \mathcal T) \, | \, A \subseteq \Omega' \}.$$

I've tried to do this in two directions. I have an ansatz for each direction:
$"\subseteq"$: We have to show that $A\in \mathcal B(\Omega, \mathcal T)$, i.e. for all $\sigma$-algebras $\mathcal A$ over $\Omega$ with $\mathcal T \subseteq \mathcal A$ it holds that $A\in \mathcal A$. So let $\mathcal A$ be an arbitrary $\sigma$-algebra with $\mathcal T \subseteq \mathcal A$. Because $\Omega' \in \mathcal B(\Omega, \mathcal T)$ we have that $\Omega' \in \mathcal A$.
Since $A \in \mathcal B(\Omega', \mathcal T')$ ...

and here's basically where I'm stuck. I don't know how to make the connection between
$\mathcal A$ being a $\sigma$-algebra over $\Omega$ and "shrinking" it to one over just $\Omega'$. Same problem in the other direction. I feel like I need to use the definition of the subspace topology but don't really know how.
Appreciate any help!

Answer 1

For any subset $\Omega'$ it's fairly easy to see that

$$\mathcal B(\Omega', \mathcal T') = \{ A \cap \Omega': A \in \mathcal B(\Omega, \mathcal T)\}$$

Proof: the right hand side is clearly a $\sigma$-algebra on $\Omega'$ and it contains all of $\mathcal{T}'$ as all subspace open sets are of the form $O \cap \Omega'$ with $O \in \mathcal{T} \subset \mathcal B(\Omega, \mathcal T)$, so by minimality we have the left to right inclusion. On the other hand, $i: \Omega' \rightarrow \Omega ,i(x) = x$ is continuous so for any Borel set $A$ of $\Omega$, $i^{-1}[A] = \Omega' \cap A$ is Borel in $\Omega'$ so the reverse inclusion also holds.

As an alternative to the last inclusion function argument: let $\mathcal{S}$ be the set of all subsets $A$ of $\Omega$ such that $A \cap \Omega'$ is Borel in $\Omega'$. One checks easily that $\mathcal{S}$ is a $\sigma$-algebra, and it contains $\mathcal{T}$by the definition of a subspace topology. So $\mathcal{B}(\Omega, \mathcal{T}) \subset \mathcal{S}$ by minimality ,and this means exactly that the right to left inclusion holds, if you think about it.

From this general fact, yours follows (we now use $\Omega' \in \mathcal B(\Omega, \mathcal T)$, because then:

$$\{ A \cap \Omega': A \in \mathcal B(\Omega, \mathcal T)\} = \{A \in \mathcal B(\Omega, \mathcal T): A \subset \Omega' \}$$

(left to right: $A \cap \Omega'$ is indeed a subset of $\Omega'$ that is Borel as $\Omega'$ is and $A$ is. right to left is trivial as $A \subset \Omega'$ means that $A \cap \Omega' = A$, so the same $A$ works.)

Answer 2

First of all, prove that $\{A\in\mathcal B(\Omega,\mathcal T)\mid A\subseteq\Omega'\}$ is in fact a $\sigma$-algebra. This is not hard at all.

Secondly, show that every open set in $\cal T'$ lies in that $\sigma$-algebra, and that it is in fact Borel in $\cal T$. Together, this would give you the needed equality.