Does $\mathcal B([0,1])=\{U\cap [0,1]\mid U\in \mathcal B(\mathbb R)\}$? 
Endow the subset $X$ of $\mathbb{R}$ with the subspace topology. Denote $\mathcal B(X)$ as the set of all the Borel sets of $X$. Does one have
  $$
\mathcal B([0,1])\underset{(1)}{=}\{U\cap [0,1]\mid U\in \mathcal B(\mathbb R)\}\ ?
$$


For (1) I tried as follow. We have that $$\{U\cap [0,1]\mid U\in \mathcal B(\mathbb R)\},$$
is a $\sigma -$algebra (easy). Since open set of $[0,1]$ are of the form $[0,1]\cap O$ where $O$ is open in $\mathbb R$, we have that $$\mathcal B([0,1])\subset \{U\cap [0,1]\mid U\in \mathcal B(\mathbb R)\}.$$
How can I prove the converse inclusion ?
 A: There is a general result:

$\textbf{Theorem}$  Let   $\mathcal{F}$ be a collection  of subsets of $X$ and $Y \subseteq X$ non-empty.Denote $$\sigma(\mathcal{F})_Y=\{A \cap Y:A \in \sigma(\mathcal{F})\}$$ $$\mathcal{F}_Y=\{A \cap Y:A \in \mathcal{F}\}$$Then $$\sigma(\mathcal{F})_Y=\sigma(\mathcal{F}_Y)$$ where $\sigma(B)$ is the sigma algebra generated by the collection of sets  $B$

$\textbf{Proof}$
Let $S \in \mathcal{F}_Y$ then $S=A \cap Y$ for some $A \in \mathcal{F} \subseteq \sigma(\mathcal{F})\Longrightarrow S \in \sigma(\mathcal{F})_Y$
and since $\sigma(\mathcal{F})_Y$ is a sigma algebra(easy exercise for you) we have that $$\sigma(\mathcal{F}_Y) \subseteq \sigma(\mathcal{F})_Y$$
Now  define the collection $\Sigma=\{A \subseteq X:A \cap Y \in \sigma(\mathcal{F}_Y)\}$
We have that $\Sigma$ is a sigma algebra of subsets of $X$(again an easy exercise for you).
If $A \in \mathcal{F}$ then $A \cap Y\in \mathcal{F}_Y \subseteq\sigma(\mathcal{F}_Y)$
hence $A \in \Sigma$
Therefore $\mathcal{F} \subseteq \Sigma\Longrightarrow \sigma(\mathcal{F}) \subseteq \Sigma$
Now for an arbitrary $B \in \sigma(\mathcal{F})_Y$ we have that $B=A \cap Y$ for some $A \in \sigma(\mathcal{F}) \subseteq \Sigma$ and thus $B \in \sigma(\mathcal{F}_Y)$
This implies that $\sigma(\mathcal{F})_Y \subseteq \sigma(\mathcal{F}_Y)$

Now apply this theorem with $\mathcal{F}=$open sets(topology of $\Bbb{R}$) and $Y=[0,1]$ and $X=\Bbb{R}$

Note that $B[0,1]=\{A \subseteq [0,1]:A \in B(\Bbb{R})\}$ and $[0,1]$ has the subspace topology.
